Question

What is the value of $E_{\text {cell }}$ ? $Cr \left| Cr ^{3+}(0.1 M )\right|\left| Fe ^{2+}(0.01\, M )\right| Fe$ Given, $E_{ Cr ^{3+} / Cr }^{\circ}=-0.74\, V$ and $E_{ Fe ^{2+} / Fe }=-0.44\, V$

• A. + 0.2941 V
• B. +0.5212 V
• C. +0.1308 V
• D. -0.2606 V

Answer 1

Answer: (A)

$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \left[\frac{\text { product }}{\text { reactant }}\right]$
Given, $E_{ Cr ^{3+} / Cr }^{\circ}=-0.74\, V$
$E_{ Fe ^{2+} / Fe }=-0.44\, V$
$Cr \left| Cr ^{3+}(0.1\, M )\right|\left| Fe ^{2+}(0.1\, M )\right| Fe$
$\therefore Cr ^{3+} / Cr$ is anode and $Fe ^{2+} / Fe$ is cathode.
$E_{\text {cell }}^{\circ} =E_{C}^{\circ}-E_{A}^{\circ}$
$=(-0.44)-(0.74)$
$=-0.44+0.74$
$=+0.30\, V$
Cell reaction is
$2 Cr +3 Fe ^{2+} \longrightarrow 2 Cr ^{3+}+3 Fe$
number of electrons in cell reaction $=6$
$E_{\text {cell }}^{\circ}=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \left[\frac{\text { product }}{\text { reactant }}\right]$
$=+0.30\, V -\frac{0.059}{6} \log \left[\frac{\left( Cr ^{3+}\right)^{2}}{\left( Fe ^{2+}\right)^{3}}\right]$
$=0.30-\frac{0.059}{6} \log \left[\frac{(0.1)^{2}}{(0.01)^{3}}\right]$
$=0.30-\frac{0.059}{6} \log 10^{4}$
$=0.30-\frac{0.059}{6} \times 0.60$
$=0.30-5.9 \times 10^{-3}$
$=0.2941\, V$