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The emf of a cell containing sodium/ copper electrodes is 3.05 V, if the electrode potential copper electrode is +0.34 V, the electrode potential of sodium is:    
  • a)
    -2.71 V        
  • b)
    +2.71 V        
  • c)
    -3.71 V        
  • d)
    +3.71 V
Correct answer is option 'A'. Can you explain this answer?
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The emf of a cell containing sodium/ copper electrodes is 3.05 V, if t...
Given Information:
- emf of the cell = 3.05 V
- electrode potential of copper electrode = 0.34 V

To Find:
- Electrode potential of sodium

Solution:

The electrode potential of a cell can be determined using the formula:

emf of the cell = electrode potential of the cathode - electrode potential of the anode

In this case, the copper electrode is the cathode (positive terminal) and the sodium electrode is the anode (negative terminal).

Given:
emf of the cell = 3.05 V
electrode potential of the copper electrode = 0.34 V

Substituting these values into the formula, we get:

3.05 V = electrode potential of copper electrode - electrode potential of sodium electrode

To find the electrode potential of sodium, we rearrange the equation:

electrode potential of sodium electrode = electrode potential of copper electrode - emf of the cell

Plugging in the given values, we have:

electrode potential of sodium electrode = 0.34 V - 3.05 V

Simplifying the expression, we get:

electrode potential of sodium electrode = -2.71 V

Therefore, the electrode potential of sodium is -2.71 V.

Answer:
The electrode potential of sodium is -2.71 V. (Option A)
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The emf of a cell containing sodium/ copper electrodes is 3.05 V, if the electrode potential copper electrode is +0.34 V, the electrode potential of sodium is: a)-2.71 V b)+2.71 V c)-3.71 V d)+3.71 VCorrect answer is option 'A'. Can you explain this answer?
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