Question

E.M.F. of the cell reaction, $2A{g}^{+}+Cu⟶2Ag+C{u}^{2+}$ is 0.46 V. If $E_{{Cu}^{2+}/Cu}$ is 0.34 V, $E_{{Ag}^{+}/Ag}$ is:

• A. 0.80 V
• B. 0.12 V
• C. 0.40 V
• D. 1.60 V

Answer 1

Answer: (C)

0.40 V

${2Ag}^{+}+Cu⟶2Ag+{Cu}^{2+},E_{r\times n}=0.46V$

$E_{{Cu}^{2+}/Cu}=0.34V\Rightarrow {Cu}^{2+}+2e^{-}\to Cu⟶\left(1\right)$

let $E_{{Ag}^{+}/Ag}=xV\Rightarrow {Ag}^{+}+e^{-}\to Ag⟶\left(2\right)$

$2\left(2\right)-\left(1\right)$

${2Ag}^{+}+{2e}^{-}+Cu⟶{Cu}^{2+}+2e^{-}+2Ag$

${2Ag}^{+}+Cu\to 2Ag+{Cu}^{2+}{E}_{cell}=2\left(x\right)-\left(0.34\right)$

But we know $E_{r\times n}=0.46V$

$0.46=2x-0.34$

$0.80=2x$

$x=0.40$

$\Rightarrow E_{{Ag}^{+}/Ag}$ is $0.40$ Volts.