At temperature of 298K, the emf of the following cell.
$A g (S) A g^{+} (0.1 M) | Z n (s)$ will be (Given, $E_{c e l l}^{0} = - 1.562 V$ )

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Solution

E = $E^{○}$ - $\frac{0.059}{n}$ log [K] at 25 degree Celsius .
n : no of electron involved
K : equilibrium constant ( not considering active mass of solid)
Ag $\to$ ${A g}^{+}$ + $e^{-}$
n = 1
Concentration of Silver ion = 0.1
Hence K = 0.1 ( we do not take active mass of solid )
E = -1.562 - 0.059 log 0.1
E= -1.503

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Q. At temperature of 298K, the emf of the following electrochemical cell will be:

A g (s) | A g^{+} (0.1 M) | | Z n^{2 +} (0.1 M) | Z n (s)

(Given,

E_{c e l l}^{o} = - 1.562 V

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At temperature of 298K, the emf of the following cell.Ag(S)Ag+(0.1M)|Zn(s) will be (Given, E0cell=−1.562V)

E = E○ - 0.059n log [K] at 25 degree Celsius .n : no of electron involved K : equilibrium constant ( not considering active mass of solid)Ag → Ag+ + e− n = 1Concentration of Silver ion = 0.1 Hence K = 0.1 ( we do not take active mass of solid )E = -1.562 - 0.059 log 0.1 E= -1.503

At temperature of 298K, the emf of the following cell.
$A g (S) A g^{+} (0.1 M) | Z n (s)$ will be (Given, $E_{c e l l}^{0} = - 1.562 V$ )