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(C) The cell reaction is: $3HSnO_2^- + Bi_2O_3 + 3H_2O + 6OH^- \rightarrow 3Sn(OH)_6^{2-} + 2Bi$.The number of electrons involved in the balanced equa

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The cell reaction is: $3HSnO_2^- + Bi_2O_3 + 3H_2O + 6OH^- ightarrow 3Sn(OH)_6^{2-} + 2Bi$.The number of electrons involved in the balanced equation is $n = 6$.The standard cell potential is $E^0_{	ext{cell}} = E^0_{	ext{cathode}} - E^0_{	ext{anode}} = -0.44 - (-0.90) = +0.46 \ V$.Applying the Nernst equation at $298 \ K$: $E_{	ext{cell}} = E^0_{	ext{cell}} - \frac{0.0591}{n} \log Q$.Using $0.06$ as an approximation for $\frac{0.0591}{1}$: $E_{	ext{cell}} = 0.46 - \frac{0.06}{6} \log(10^6)$.$E_{	ext{cell}} = 0.46 - 0.01 	imes 6 = 0.46 - 0.06 = 0.40 \ V$.Thus, $E_{	ext{cell}} = 4 	imes 10^{-1} \ V$.The value is $4$.

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