Question

The measured voltage for the reaction with the indicated concentrations is $1.5\text{V}$. Calculate $E_{\text{cell}}^{\circ }\text{in V}$.
$Cr\left(s\right)+3A{g}^{+}(aq,0.1\text{M})\to 3Ag\left(s\right)+C{r}^{3+}(aq,0.1\text{M})$

• A. $1.35$
• B. $1.46$
• C. $1.65$
• D. $1.54$

Answer 1

The correct option is D $1.54$

$Cr\left(s\right)+3A{g}^{+}(aq,0.1\text{M})\to 3Ag\left(s\right)+C{r}^{3+}(aq,0.1\text{M})$
Here number of electrons involved $=3$
Here $E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.0591}{3}\log \frac{\left[C{r}^{3+}\right]}{[A{g}^{+}{]}^3}\Rightarrow E_{\text{cell}}^{\circ }=1.5+\frac{0.0591}{3}\log \frac{0.1}{{10}^{-3}}\Rightarrow E_{\text{cell}}^{\circ }=1.5+\frac{0.0591}{3}\log 100\Rightarrow E_{\text{cell}}^{\circ }=1.5+\frac{0.0591}{3}\times 2\Rightarrow E_{\text{cell}}^{\circ }=1.54\text{V}$