Sn4+ (1.50 M) +Zn (s) ----> Sn2+ (0.50M) + Zn2+
i. Calculate the potential of the following cell reaction at 298 K.
ii.The standard potential E of the cell is 0.89 V. Whether the potential of cell will increase or decrease if concentration of Sn4+ is increased in the cell

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Solution

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${Sn}^{4 +} + Zn \to {Sn}^{2 +} + {Zn}^{2 +} At anode : Zn \to {Zn}^{2 +} + 2 e^{-} At cathode : {Sn}^{4 +} + 2 e^{-} \to {Sn}^{2 +} Here, n = 2 (i) E = E ° - \frac{0.059}{n} \log \frac{[{Sn}^{2 +}]}{[{Sn}^{4 +}]} = 0.89 + \frac{0.059}{2} \log \frac{0.50}{1.50} = 0.89 + \frac{0.059}{2} \times (- 0.4771) = 0.89 - 0.0140 = 0.87 V (ii) If the concentration of {Sn}^{4 +} will increase then the ratio of \frac{[{Sn}^{2 +}]}{[{Sn}^{4 +}]} will decrease and so the \log value of this ratio will increase and overall, the potential of the cell will decrease .$

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