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For an electrochemical cell Sn(s)Sn^(2+)...

For an electrochemical cell `Sn(s)Sn^(2+) (aq,1M)||pb^(2+) (Aq,1M)|Pb(s)` the ratio `([Sn^(2+)])/([Pb^(2+)]` when this cell attians equilibrium is_______ (Given : `E_(sn^(2+)|Sn)^(@) = -0.14V, E_(pb^(2+|Pb^(@))= -0.13V, (2.303RT)/(F = 0.06))`

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To solve the problem, we need to find the ratio \(\frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]}\) when the electrochemical cell reaches equilibrium. We will use the Nernst equation to do this. ### Step-by-Step Solution: 1. **Identify the Half-Reactions and Standard Potentials**: - The half-reaction for tin (Sn) is: \[ \text{Sn}^{2+} + 2e^- \rightarrow \text{Sn} \quad E^\circ = -0.14 \, \text{V} ...
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For an electrochemical cell Sn(s)|Sn^(2+)(aq, 1M)||Pb^(2+)(aq, 1M)|Pb(s) the ratio ([Sn^(2+)])/([Pb^(2+)]) when this cell attains equilibrium is _________________. ("Given "E_(Sn^(2+)|Sn)^(@)=-0.14V, " "E_(Pb^(2+)|Pb)^(@)=-0.13V,(2.303RT)/(F)=0.06)

Given: E_(Sn^(2+)//Sn) ^0 = -0.14V , E_(Pb^(2+)//Pb) ^0 = -0.13V . Determine [(Sn^(2+))//(Pb^(2+))] at equilibrium. For cell reaction Sn |Sn^(2+) || Pb^(2+) | Pb , Take (2.303RT)//F = 0.06V