Pt(s)|H_2(g)(1 bar)|H^ + (aq)(1 M)||M^3 + (aq),M^ + (aq)|Pt(s) The E_cell for the given cell is 0.1115 V at 298 K when [ M^ + (aq) ] / [ M^3 + (aq) ] = 10^a The value of a is ____________ Given : E_M^3 + /M^ + ^theta = 0.2 V 2.303RT / F = 0.059V

Overall reaction :- H_2( g)+ M_( aq)^3+ -> M_ (aq) ^++2 H_( aq)^+ \ E_ Cell = E_ Cathode ^ o- E_ anode ^ o-(0.059 / 2) log frac[ M^+] x 1^2[ M^+3] 1 \ 0.1115=0.2-(0.059 / 2) log frac[ M^+][ M^+3] \ 3=log frac[ M^+][ M^+3] \ Therefore a=3

PtsH2g1barH aq1MM3 aqM aqPts The Ecell for the given ce... | null (Chemistry)

Overall reaction :-

\begin{aligned} & \mathrm{H}_{2(\mathrm{~g})}+\mathrm{M}_{(\mathrm{aq})}^{3+} \longrightarrow \mathrm{M}_{\text {(aq) }}^{+}+2 \mathrm{H}_{(\mathrm{aq})}^{+} \\\\ & \mathrm{E}_{\text {Cell }}=\mathrm{E}_{\text {Cathode }}^{\mathrm{o}}-\mathrm{E}_{\text {anode }}^{\mathrm{o}}-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right] \times 1^2}{\left[\mathrm{M}^{+3}\right] 1} \\\\ & 0.1115=0.2-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\\\ & 3=\log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\\\ & \therefore \mathrm{a}=3 \end{aligned}

Explanation

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