1. Tardigrade
  2. Question
  3. Chemistry
  4. For the cell C u(s)|C u2+(a q)(0.1 M) | A g+(a q)(0.01 M)| A g(s) the cell potential E1=0.3095 V For the cell C u(s)|C u2+(a q)(0.01 M) | A g+(a q)(0.001 M)| A g(s) the cell potential =10-2 V. (Round off the Nearest Integer). [Use: (2.303 R T/F)=0.059]

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Q. For the cell
$C u(s)\left|C u^{2+}(a q)(0.1 M) \| A g^{+}(a q)(0.01 M)\right| A g(s)$
the cell potential $E_{1}=0.3095 V$ For the cell
$C u(s)\left|C u^{2+}(a q)(0.01 M) \| A g^{+}(a q)(0.001 M)\right| A g(s)$
the cell potential =_____$10^{-2} V$.
(Round off the Nearest Integer).
[Use : $\frac{2.303 R T}{F}=0.059$]

JEE MainJEE Main 2021Electrochemistry

Solution:

Cell reaction is :
$C u(s)+2 A g^{+}(a q) \rightarrow C u^{2+}(a q)+2 A g(s)$
Now, $E_{\text {cell }}=E_{\text {Cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[ Cu ^{2+}\right]}{\left[ Ag ^{+}\right]^{2}} ......(1)$
$\therefore E_{1}=0.3095=E_{\text {Cell }}^{\circ}-\frac{0.059}{2} \cdot \log \frac{0.01}{(0.001)^{2}}$
From $(1)$ and $(2), E_{2}=0.28 V =28 \times 10^{-2} V$