Two elements A and B have reduction potential’s 0.23V and 0.46V ...
The more positive the reduction potential value, more is the tendency of the element to reduce itself and hence more will be the tendency to accept electrons. Hence the other element will have more tendency to lose electrons.
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Two elements A and B have reduction potential’s 0.23V and 0.46V ...
S of -0.5 V and -0.2 V, respectively. This means that element A is more easily oxidized (loses electrons) than element B. In other words, element A has a greater tendency to donate electrons and become a cation than element B.
If the two elements were placed in a cell with an electrode potential of -0.3 V (between the reduction potentials of A and B), element A would be oxidized and element B would be reduced. Electrons would flow from element A to element B, creating a current. The overall reaction would be:
A (oxidized) + B+ (reduced) → A+ (reduced) + B (oxidized)
The cell potential can be calculated using the formula Ecell = Ecathode - Eanode, where Ecathode is the reduction potential of the cathode (B in this case) and Eanode is the reduction potential of the anode (A in this case). In this scenario, the cell potential would be:
Ecell = -0.2 V - (-0.5 V) = 0.3 V
This positive cell potential indicates that the reaction is spontaneous and that the flow of electrons is favorable.
Two elements A and B have reduction potential’s 0.23V and 0.46V ...
The more positive the reduction potential value, more is the tendency of the element to reduce itself and hence more will be the tendency to accept electrons. Hence the other element will have more tendency to lose electrons