The Question : Represent the cell in which the following reaction takes place. The value of E˚ for the cell is 1.260 V. What is the value of Ecell? 2Al(s) + 3Cd2+ (0.1M) —-> 3Cd (s) + 2Al3+ (0.01M)
Solution for the question :
Al(s) /Cd2+ (0.1M) // Al3+ (0.01M) /Cd(s) 2Al(s) + 3Cd2+ (0.1M) —> 3Cd (s) + 2Al3+ (0.01M) Ecell = Eocell -0.059/n log [Al3+]2 / [Cd2+]3 Ecell= 1.26 – 0.059/6 log (0.01)2 /(0.1)3 = 1.26 – 0.059 (-1) /6 = 1.26+0.009 = 1.269 V
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