Question

Consider an electrochemical cell in which the following reaction occurs and predict which changes will decreases the cell voltage:

$F{e}^{2+}\left(aq\right)+A{g}^{+}\left(aq\right)\to Ag\left(s\right)+F{e}^{3+}\left(aq\right)$

(I) decreases the $\left[A{g}^{+}\right]$ (II) increases in $\left[F{e}^{3+}\right]$ (III) increase the amount of $Ag$

• A. II and III only
• B. II only
• C. I and II only
• D. I only

Answer 1

Answer: (C)

The correct option is D I and II only
$F{e}_{\left(aq\right)}^{2+}+A{g}_{\left(aq\right)}^{+}\to A{g}_{\left(s\right)}+F{e}_{\left(aq\right)}^{3+}$

$E_{cell}=E_{cell}^o-\frac{0.0591}{1}log\frac{\left[F{e}^{3+}\right]}{\left[F{e}^{2+}\right]\left[A{g}^{+}\right]}$

(I) On increasing $\left[F{e}^{3+}\right]$
$-log\frac{\left[F{e}^{3+}\right]}{\left[F{e}^{2+}\right]\left[A{g}^{+}\right]}$ decreases

$\because E_{Cell}^o$ is constant.

So, $E_{cell}$ will decrease

(II) On decreasing $\left[A{g}^{+}\right]$
$-log\frac{\left[F{e}^{3+}\right]}{\left[F{e}^{2+}\right]\left[A{g}^{+}\right]}$ decreases

So, $E_{cell}$ will decrease

(III) Changing amount of $Ag\left(s\right)$ will not affect $E_{cell}$

Option D is correct.