MCQ
Given $E^o _{Cr^{3+} / Cr} = - 0.72\, V, \,$$E^o _{Fe^{2+} / Fe} =-0.42 \,V.$ The potential for the cell
$Cr\, | \,Cr^{3+}_{(0.1\,M)}\,||\, Fe^{2+}_{(0.01\, M)}\,|\, Fe$ is ......... $V$
- A$-0.26$
- B$0.336$
- C$-0.339$
- ✓$0.26$
✓
Answer
Correct option: D.
$0.26$
d
$E_{c e l l}=E_{F e^{2+} / F e}^{0}-E_{C r^{3}+/ C r}^{\circ}-$$\frac{0.059}{6} \log \frac{\left[C r^{3+}\right]^{2}}{\left[F e^{2 t}\right]^{3}}$
$E_{c e l l}=E_{F e^{2+} / F e}^{0}-E_{C r^{3}+/ C r}^{\circ}-$$\frac{0.059}{6} \log \frac{\left[C r^{3+}\right]^{2}}{\left[F e^{2 t}\right]^{3}}$
$=-0.42-(-0.72)-\frac{0.059}{6} \log \frac{(0.1)^{2}}{(0.01)^{3}}$
$=-0.42+0.72-\frac{0.059}{6} \log \frac{0.1 \times 0.1}{0.01 \times 0.01 \times 0.01}$
$=0.3-\frac{0.059}{6} \log \frac{10^{-2}}{10^{-6}}$
$=0.3-\frac{0.059}{6} \times 4$
$=0.30-0.0393=0.26 V$
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