Considering the reaction Cl_{2(g)} + 2Br^{-}_ | vedclass

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(B) The Nernst equation for the reaction is: $E_{cell} = E^o_{cell} - \frac{0.059}{n} \log \frac{[Cl^{-}]^2 [Br_2]}{P_{Cl_2} [Br^{-}]^2}$ Here,$n = 2$

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$0.35$

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(B) The Nernst equation for the reaction is: $E_{cell} = E^o_{cell} - \frac{0.059}{n} \log \frac{[Cl^{-}]^2 [Br_2]}{P_{Cl_2} [Br^{-}]^2}$ Here,$n = 2$,$E^o_{cell} = 0.29 \ V$,$[Cl^{-}] = 0.01 \ M$,$[Br_2] = 0.01 \ M$,$[Br^{-}] = 0.01 \ M$,and $P_{Cl_2} = 1 \ atm$. Substituting the values: $E_{cell} = 0.29 - \frac{0.059}{2} \log \frac{(0.01)^2 (0.01)}{1 	imes (0.01)^2}$ $E_{cell} = 0.29 - 0.0295 \log(0.01)$ $E_{cell} = 0.29 - 0.0295 	imes (-2) = 0.29 + 0.059 = 0.349 \ V \approx 0.35 \ V$.

Considering the reaction $Cl_{2(g)} + 2Br^{-}_{(aq)} \rightarrow 2Cl^{-}_{(aq)} + Br_{2(l)}$,calculate the cell $emf$ in $V$ when $[Cl^{-}] = [Br_2] = [Br^{-}] = 0.01 \ M$ and $Cl_2$ gas is at $1 \ atm$ pressure. (Given $E^o = 0.29 \ V$ for the reaction)

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