Page 1 CBSE TEST PAPER-02 CLASS - XII CHEMISTRY (Electrochemistry) [ANSWERS] Topic: -Nernst equation and its applications 1. E 0 cell =E 0 cathode – E 0 Anode =E 0 Ag +/Ag – E 0 Cu 2+ /Cu =0.80V – (+0.34V) =+0.46V 2. From the reduction potential values, it is indicated that Nickel (more negative value) is more reactive than copper and will, then displace copper from CuSO 4 Ni (s) + Cu 2+ (aq) ----Ni 2 + (aq) +Cu(s). 3. Factors affecting electrode potential values are – a) Concentration of electrolyte b) Temperature. 4. Daniel cell: Zn(s) / Zn 2 +(aq) //Cu 2 +(aq)/Cu(s) Nernst equation – at 298 K E cell = (E 0 Cu 2 +/Cu – E 0 Zn 2 +/Zn) - 2 2 0.059 [ ] log 2 [ ] Zn cu + + 5. (i) Standard electrode potential and equilibrium constant E 0 cell = c 2.303 log k RT nF Where E 0 cell = standard electrode potential of cell R = Gas constant T = temperature in Kelvin n = no .of electrons. F = Faraday’s constant and Kc = Equilibrium constant (ii) Standard electrode potential and Gibbs free energy change- ? G 0 = - n F E 0 cell Where ? G 0 = Change in Gibbs’ free energy n = No. of electrons Page 2 CBSE TEST PAPER-02 CLASS - XII CHEMISTRY (Electrochemistry) [ANSWERS] Topic: -Nernst equation and its applications 1. E 0 cell =E 0 cathode – E 0 Anode =E 0 Ag +/Ag – E 0 Cu 2+ /Cu =0.80V – (+0.34V) =+0.46V 2. From the reduction potential values, it is indicated that Nickel (more negative value) is more reactive than copper and will, then displace copper from CuSO 4 Ni (s) + Cu 2+ (aq) ----Ni 2 + (aq) +Cu(s). 3. Factors affecting electrode potential values are – a) Concentration of electrolyte b) Temperature. 4. Daniel cell: Zn(s) / Zn 2 +(aq) //Cu 2 +(aq)/Cu(s) Nernst equation – at 298 K E cell = (E 0 Cu 2 +/Cu – E 0 Zn 2 +/Zn) - 2 2 0.059 [ ] log 2 [ ] Zn cu + + 5. (i) Standard electrode potential and equilibrium constant E 0 cell = c 2.303 log k RT nF Where E 0 cell = standard electrode potential of cell R = Gas constant T = temperature in Kelvin n = no .of electrons. F = Faraday’s constant and Kc = Equilibrium constant (ii) Standard electrode potential and Gibbs free energy change- ? G 0 = - n F E 0 cell Where ? G 0 = Change in Gibbs’ free energy n = No. of electrons F = Faraday’s Constant E 0 cell = Standard electrode Potential of cell. 6. Fe 3+ + 3e - Fe According to Nernst Equation – E Fe 3+ / Fe = E 0 Fe 3+ /Fe – 0.059 n log 3 1 [ ] Fe + = 0.771 V - 0.059 1 log 3 0.1 = 0.771 V - 0.0197 V = +0.7513V 7. P H = - Log [H + ] The cell reaction is – H + + e- 1 2 H 2 (g) According to Nearest Equation E = E 0 - 0.059 1 log [ ] n H + 0.03V = 0 + 0.059 1 (- 1 log [ ] H + ) = 0 + 0.059 P H PH = 0.03 0.059 V = 5.07 V 8. The cell reaction is 2 Cr + 3 Fe 2+ 6e - ? 2 Cr 3+ + 3Fe Nernst Equation – E cell = ( E 0 Fe 2 +/ Fe- E 0 cr 3+ /cr ) - 2 3 3 2 0.059 log 6 Cr Fe + + ? ? ? ? ? ? ? ? =(-0.44v – (-0.74v) - ( ) ( ) 2 3 0.10 0.059 log 6 0.01 = 0.3V - 4 0.059 log 10 6 = 0.3V – 0.0394V = +0.2606 V Page 3 CBSE TEST PAPER-02 CLASS - XII CHEMISTRY (Electrochemistry) [ANSWERS] Topic: -Nernst equation and its applications 1. E 0 cell =E 0 cathode – E 0 Anode =E 0 Ag +/Ag – E 0 Cu 2+ /Cu =0.80V – (+0.34V) =+0.46V 2. From the reduction potential values, it is indicated that Nickel (more negative value) is more reactive than copper and will, then displace copper from CuSO 4 Ni (s) + Cu 2+ (aq) ----Ni 2 + (aq) +Cu(s). 3. Factors affecting electrode potential values are – a) Concentration of electrolyte b) Temperature. 4. Daniel cell: Zn(s) / Zn 2 +(aq) //Cu 2 +(aq)/Cu(s) Nernst equation – at 298 K E cell = (E 0 Cu 2 +/Cu – E 0 Zn 2 +/Zn) - 2 2 0.059 [ ] log 2 [ ] Zn cu + + 5. (i) Standard electrode potential and equilibrium constant E 0 cell = c 2.303 log k RT nF Where E 0 cell = standard electrode potential of cell R = Gas constant T = temperature in Kelvin n = no .of electrons. F = Faraday’s constant and Kc = Equilibrium constant (ii) Standard electrode potential and Gibbs free energy change- ? G 0 = - n F E 0 cell Where ? G 0 = Change in Gibbs’ free energy n = No. of electrons F = Faraday’s Constant E 0 cell = Standard electrode Potential of cell. 6. Fe 3+ + 3e - Fe According to Nernst Equation – E Fe 3+ / Fe = E 0 Fe 3+ /Fe – 0.059 n log 3 1 [ ] Fe + = 0.771 V - 0.059 1 log 3 0.1 = 0.771 V - 0.0197 V = +0.7513V 7. P H = - Log [H + ] The cell reaction is – H + + e- 1 2 H 2 (g) According to Nearest Equation E = E 0 - 0.059 1 log [ ] n H + 0.03V = 0 + 0.059 1 (- 1 log [ ] H + ) = 0 + 0.059 P H PH = 0.03 0.059 V = 5.07 V 8. The cell reaction is 2 Cr + 3 Fe 2+ 6e - ? 2 Cr 3+ + 3Fe Nernst Equation – E cell = ( E 0 Fe 2 +/ Fe- E 0 cr 3+ /cr ) - 2 3 3 2 0.059 log 6 Cr Fe + + ? ? ? ? ? ? ? ? =(-0.44v – (-0.74v) - ( ) ( ) 2 3 0.10 0.059 log 6 0.01 = 0.3V - 4 0.059 log 10 6 = 0.3V – 0.0394V = +0.2606 V 9. The half cell reactions are Anode : Zn(s) ? Zn 2 + (aq) +2e - Cathode : Cd 2 + (aq) + 2e - ? Cd (s) Nernst Equation E cell = (E 0 Cathode - E 0 anode) - 2 2 0.059 log Zn n Cd + + ? ? ? ? ? ? ? ? = (- 0.403 – (-0.763) - 0.059 0.0004 log 2 0.2 = 0.36V - 0.0798V = 0.4398V 0 0 cell = - n F E G ? = -2mol × 96500 C/mol × 0.4398V = -8488 J mol -1 10. From the reaction, n =2 E 0 cell = E 0 cu 2 +/cu - E 0 Zn 2 +/Zn = + 0.34v - (-0.76v) = 1.10V E 0 cell = c 2.303 log k RT nF At 298k , E 0 cell × c log k 0.059 n Log k c = E 0 cell × 0.059 n = 1.10 × 2 = 37.29 0.059 K c = Antilog 37.29 = 1.95 ×10 37 11. [Ag + ] = 5.3 ×10 -9 M 12. E 0 cell = 0.03V 0 = -2895J G ?Read More

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