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The standard reduction potential of $Cu^{+2}/Cu$ and $Ag^+/Ag$ electrode are 0.337 and 0.799Volt respectively for this positive emf galvanic cell,for what concentration of $Ag^+$ will be emf of cell at $25^{\large\circ}C$ be zero,if the concentration of $Cu^{+2}$ is 0.01M?

$\begin{array}{1 1}(a)\;1.37\times 10^9M\\(b)\;1.67\times 10^9M\\(c)\;1.57\times 10^9M\\(d)\;1.47\times 10^9M\end{array}$

1 Answer

Since $E^0_{Ag^+/Ag}=0.799V$ is greater than $E^0_{Cu^{+2}/cu}=0.337V$
$Ag^+(aq)/Ag(s)$ will form the cathode end $Cu(s)/Cu^{+2}(aq)$ will behave as anode of galvanic cell.Thus,the cell can be represented as given below.
Here $M_1$ is concentration of $Ag^+(aq)$ ions.
$Cu(s)/Cu^{+2}(0.001M)\parallel Ag^+(c)/Ag(s)$
Cell reaction is $Cu(s)+2Ag^+(aq)\rightarrow Cu^{+2}(aq)+2Ag(s)$
Here $n=2$
From Nernst equation at 298K,we have
$E_{cell}=E^0_{cell}-\large\frac{0.0591}{2}$$\log \large\frac{[Cu^+(aq)]}{[Cu(s)]}\frac{[Ag(s)]}{[Ag^+(aq)]^2}$
$\Rightarrow [E^0_{Ag^+/Ag}-E^0_{Cu^{+2}/Cu}]-0.0295\log \large\frac{0.01\times 1}{1\times C^2}$
At equlibrium $E_{cell}=0$
$0=(0.799-0.337)-0.0295\log \large\frac{0.01}{C^2}$
$0.462=0.0295\log \large\frac{0.01}{C^2}$
$\log \large\frac{0.01}{C^2}=\frac{0.462}{0.0295}$$=15.6610$
$\log \large\frac{0.01}{C^2}$$=antilog 15.6610=4.571\times 10^{15}$
$C^2=\large\frac{0.01}{4.57\times 10^{15}}=$$2.1877\times 10^{-18}$
Hence $[Ag^+]=1.47\times 10^9M$
Hence (d) is the correct answer.
answered Feb 17, 2014 by sreemathi.v

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