Which of the following is correct as a Nernst | vedclass

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(D) The cell reaction is as follows:Anode (Oxidation): $Mg_{(s)} \rightarrow Mg_{(aq)}^{2+} + 2e^-$Cathode (Reduction): $Cl_{2_{(g)}} + 2e^- \rightarr

Vedclass · Accepted Answer

$E_{cell} = E_{cell}^0 - \frac{0.059}{2} \log [Mg^{2+}][Cl^{-}]^2$

Vedclass · Answer

(D) The cell reaction is as follows:Anode (Oxidation): $Mg_{(s)} \rightarrow Mg_{(aq)}^{2+} + 2e^-$Cathode (Reduction): $Cl_{2_{(g)}} + 2e^- \rightarrow 2Cl_{(aq)}^{-}$Overall reaction: $Mg_{(s)} + Cl_{2_{(g)}} \rightarrow Mg_{(aq)}^{2+} + 2Cl_{(aq)}^{-}$The Nernst equation is given by: $E_{cell} = E_{cell}^0 - \frac{0.059}{n} \log Q$Here,$n = 2$ and the reaction quotient $Q = \frac{[Mg^{2+}][Cl^{-}]^2}{P_{Cl_2}}$.Since $P_{Cl_2} = 1 \ bar$,$Q = [Mg^{2+}][Cl^{-}]^2$.Thus,the correct equation is $E_{cell} = E_{cell}^0 - \frac{0.059}{2} \log [Mg^{2+}][Cl^{-}]^2$.

Which of the following is correct as a Nernst equation for the given electrochemical cell ?
$Mg_{(s)}|Mg_{(aq)}^{2+}(0.1 \ M)||Cl_{(aq)}^{-}(0.1 \ M)|Cl_{2_{(g)}}(1 \ bar)|Pt_{(s)}$

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Which of the following is correct as a Nernst equation for the given electrochemical cell ?$Mg_{(s)}|Mg_{(aq)}^{2+}(0.1 \ M)||Cl_{(aq)}^{-}(0.1 \ M)|Cl_{2_{(g)}}(1 \ bar)|Pt_{(s)}$