The e-m-f- of the cell at 25- CCu - Cu 2-0-01 M Ag -0-1 M - Agis - E 0 Cu 2- Cu -0-34 V- and -E 0 Ag - Ag -0-80 V -A- 0-46 VB- 1-14 VC- 0-43 VD- 1-11 V

The correct option is A 0.46V The cell reaction is Cu(s) + 2Ag^2+→ Cu^2+ + 2Ag(s) = E^0 0.059/2 log ( Cu^2+/Cu ) E^0 = E^0( Ag^+/Ag) E^0( Cu^2/Cu) ...

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Byju's Answer

Standard X

Chemistry

Application of Electrolysis

The e.m.f. of...

Question

The e.m.f. of the cell at $25^{\circ} C$
$\frac{C u}{C u^{2 +} (0.01 M)} | | \frac{A g^{+} (0.1 M)}{A g}$
is $[E^{0} (\frac{C u^{2 +}}{C u}) = 0.34 V a n d E^{0} (\frac{A g +}{A g}) = 0.80 V]$

0.46V

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1.14V

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0.43V

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1.11V

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Solution

The correct option is A
0.46V

$T h e c e l l r e a c t i o n i s$

$C u (s) + 2 A g^{2 +} \to C u^{2 +} + 2 A g (s)$

$= E^{0} - \frac{- 0.059}{2} l o g (\frac{C u^{2 +}}{C u})$

$E^{0} = E^{0} (\frac{A g^{+}}{A g}) - E^{0} (\frac{C u^{2}}{C u}) = 0.80 - 0.34 = 0.46 V$

$N o w, E = 0.46 -$ $\frac{0.059}{2} l o g (\frac{0.01}{(0.1)^{2}})$

$= 0.46 - \frac{0.059}{2} l o g 1$

$= 0.46 - 0 = 0.46 V$

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The e.m.f. of the cell at $25^{\circ} C$
$\frac{C u}{C u^{2 +} (0.01 M)} | | \frac{A g^{+} (0.1 M)}{A g}$
is $[E^{0} (\frac{C u^{2 +}}{C u}) = 0.34 V a n d E^{0} (\frac{A g +}{A g}) = 0.80 V]$