Find the emf of the following cell- Take RT ln 10-F-0-06Pt- H2g-H- aq-H- aq-H2 g- Pt 10atm 10-9M 10-3M 0-1atm

Cell is equivalent to Pt, H_2(g)/H^+ (aq)//H^+ (aq)/H_2 (g), Pt nbsp; 10atm 10^ 9M 10^ 3M 0.1atm oxidation half cell: 10 atmH_2(g)→10^ 9 M2H^++2e^ redu ...

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Byju's Answer

Standard XIII

Chemistry

Electrochemical Series

Find the emf ...

Question

Find the emf of the following cell. (Take $\frac{R T l n 10}{F} = 0.06$
$\begin{matrix} P t, H_{2} (g) / H^{+} (a q) / / H^{+} (a q) / H_{2} (g), P t 10 a t m 10^{- 9} M 10^{- 3} M 0.1 a t m \end{matrix}$

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Solution

Cell is equivalent to $\begin{matrix} P t, H_{2} (g) / H^{+} (a q) / / H^{+} (a q) / H_{2} (g), P t 10 a t m 10^{- 9} M 10^{- 3} M 0.1 a t m \end{matrix}$
oxidation half cell:
$H_{2} (g) 10 a t m \to 2 H^{+} 10^{- 9} M + 2 e^{-}$
reduction half cell:
$2 H^{+} (a q .) 10^{- 3} M + 2 e^{-} \to H_{2} (g) 0.1 a t m$
overall reaction:
$2 H^{+} (a q .) 10^{- 3} M + H_{2} (g) 10 a t m \to H_{2} (g) 0.1 a t m + 2 H^{+} 10^{- 9} M$
apply nearst equation for $n = 2$
$E_{c e l l}^{0} = 0 - \frac{0.06}{2} l o g \frac{(10^{- 9})^{2} \times 0.1}{10 \times (10^{- 3})^{2}} = - \frac{0.06}{2} l o g 10^{- 14} = 0.42 V$

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Find the emf of the following cell. (Take RT ln 10F=0.06 Pt, H2(g)/H+(aq)//H+(aq)/H2(g), Pt10atm 10−9M 10−3M 0.1atm

Find the emf of the following cell. (Take $\frac{R T l n 10}{F} = 0.06$
$\begin{matrix} P t, H_{2} (g) / H^{+} (a q) / / H^{+} (a q) / H_{2} (g), P t 10 a t m 10^{- 9} M 10^{- 3} M 0.1 a t m \end{matrix}$