Question

The equilibrium constant for the following general reaction is ${10}^{30}$. Calculate $E^0$ for the cell at 298K.
$2X_2\left(s\right)+{3Y}^{2+}\left(aq\right)⟶{2X}_2^{3+}\left(aq\right)+3Y\left(s\right)$ :

• A. $+0.105V$
• B. $+0.2955V$
• C. $0.0985V$
• D. $-0.2955V$

Answer 1

The correct option is B $+0.2955V$

$K_{eq}={10}^{30}$

$E^o=?$

$2X_2\left(s\right)+3Y^{2+}\left(aq\right)\to 2X_2^{3+}\left(aq\right)+3Y\left(s\right)$

$n=2\times 3=6$

At equation, $E_{cell}=0$

$E_{cell}=E_{cell}^o-\frac{0.0591}{n}log\frac{\left[P\right]}{\left[R\right]}$

At equation, $\frac{\left[P\right]}{\left[R\right]}=K_{eq}$

$0=E_{cell}^o-\frac{0.0591}{n}log{K}_{eq}$

$E_{cell}^o=\frac{0.0591}{n}log{K}_{eq}$

$log{K}_{eq}=\frac{n\times E_{cell}^o}{0.0591}$

$log{10}^{30}=\frac{6\times E_{cell}^o}{0.0591}$

$30log10=\frac{6\times E_{cell}^o}{0.0591}$

$E_{cell}^o=5\times 0.0591=0.2955V$