Given E∘Cr3+∣Cr=−0.72V and E∘Fe2+∣Fe=−0.42V The potential of the cell CrCr3+(0.1M)∥Fe2+(0.01M)Fe is - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$- 0.26 V$
B
$0.26 V$
C
$0.339$
D
$- 0.329$

Solution:

The cell potential is given by $E_{cell} = E_{R} - E_{L}$ where

Right half-cell ${Fe}^{2 +} + 2 e^{-} \to Fe$

$E_{R} = E_{R}^{\circ} - \frac{R T}{2 F} \ln \frac{1}{[{Fe}^{2 +}] / c^{\circ}} = - 0.42 V + (\frac{0.059 V}{2}) \log (0.01) = - 0.42 V - 0.06 V = - 0.48 V$

Left half-cell ${Cr}^{3 +} + 3 e^{-} \to Cr$

$E_{L} = E_{L}^{\circ} - \frac{R T}{3 F} \ln \frac{1}{[{Cr}^{3 +}] / c^{\circ}} = - 0.72 + (\frac{0.059 V}{3}) \log (0.1) = - 0.72 V - 0.02 V = - 0.74 V$

Hence, $E_{cell} = - 0.48 V - (- 0.74 V) = 0.26 V$

Given $E^{\circ} ({Cr}^{3 +} ∣ Cr) = - 0.72 V$ and $E^{\circ} ({Fe}^{2 +} ∣ Fe) = - 0.42 V$ The potential of the cell $Cr |{Cr}^{3 +} (0.1 M) ∥ {Fe}^{2 +} (0.01 M)| Fe$ is

Solution:

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