For an electrochemical cell, $Sn (s) / {Sn}^{2 +} (aq), 1 M) / / {Pb}^{2} + (aq, 1 M) / Pb (s)$ . The ratio $[{Sn}^{2 +}] / [{Pb}^{2 +}]$ when this cell attains equilibrium is ______.
$Given : E^{°} Sn (s) / {Sn}^{2 +} = 0.14 V; E^{°} {Pb}^{2 +} / Pb) / {Sn}^{2 +} = 0.13 V, 2.303 RT / F = 0.06 V$

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For an electrochemical cell, $Sn (s) / {Sn}^{2 +} (aq, 1 M) / / {Pb}^{2} + (aq, 1 M) / Pb (s)$ . The ratio [Sn²⁺]/[Pb²⁺] when this cell attains equilibrium is $2.15$
Step 1: $Given : E^{°} Sn (s) / {Sn}^{2 +} = 0.14 V; E^{°} {Pb}^{2 +} / Pb) / {Sn}^{2 +} = 0.13 V, 2.303 RT / F = 0.06 V$
Step 2: Half anodic reaction involved : $Sn \to {Sn}^{2 +} + 2 e^{-}$
Half Cathodic reaction involved : ${Pb}^{2 +} + 2 e^{-} \to Pb$
Net reaction: $Sn + {Pb}^{2 +} + 2 e^{-} \to Pb + {Sn}^{2 +}$
${E^{°}}_{cell} = {E^{°}}_{cathode} - {E^{°}}_{anode}$
Step 3: According to the Nernst equation, ${E^{°}}_{cell} = {E^{°}}_{cathode} - (0.06 / 2) \log [{Sn}^{2 +}] / [{Pb}^{2 +}]$
${E^{}}_{cell} = {E^{°}}_{cell} - (0.06 / 2) \log [{Sn}^{2 +}] / [{Pb}^{2 +}]$
${E^{°}}_{cell} = 0.01 V$
At equilibrium state ${E^{}}_{cell} = 0$
$0 = 0.01 - (0.06 / 2) \log [{Sn}^{2 +}] / [{Pb}^{2 +}] 0.01 = (0.06 / 2) \log [{Sn}^{2 +}] / [{Pb}^{2 +}] \log [{Sn}^{2 +}] / [{Pb}^{2 +}] = 1 / 3 [{Sn}^{2 +}] / [{Pb}^{2 +}] = 2.15$
Hence, The ratio $[{Sn}^{2 +}] / [{Pb}^{2 +}]$ when this cell attains equilibrium is $2.15$

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For an electrochemical cell, Sn(s)/Sn2+(aq),1M)//Pb2+(aq,1M)/Pb(s). The ratio [Sn2+]/[Pb2+] when this cell attains equilibrium is ______.Given:E°Sn(s)/Sn2+=0.14V;E°Pb2+/Pb)/Sn2+=0.13V,2.303RT/F=0.06V

For an electrochemical cell, $Sn (s) / {Sn}^{2 +} (aq), 1 M) / / {Pb}^{2} + (aq, 1 M) / Pb (s)$ . The ratio $[{Sn}^{2 +}] / [{Pb}^{2 +}]$ when this cell attains equilibrium is ______.
$Given : E^{°} Sn (s) / {Sn}^{2 +} = 0.14 V; E^{°} {Pb}^{2 +} / Pb) / {Sn}^{2 +} = 0.13 V, 2.303 RT / F = 0.06 V$