E cell = 1.962 V
Given
E° cell = 2.00 V
Required
E cell at 298 K
Solution
The potential cell for nonstandard conditions we can use the Nernst equation
[tex]\rm E = E ^ o -\dfrac {RT} {nF} lnQ[/tex]
For temperature T = 298 K,
[tex]\rm E = E ^ o -\dfrac {0.0592V} {n} log \: Q[/tex]
Reaction
3A(s) + B+3(aq) → 3A+(aq) + B(s)
Half Reaction
3A(s)⇒3A⁺(aq)+3e⁻
B⁺³(aq)+3e⁻⇒B(s)
n = 3 (3 electron transfer)
Q = the reaction quotient :
[tex]\tt Q=\dfrac{[A^+]^3}{[B^{+3}]}=\dfrac{0.5^3}{1.5.10^{-3}}=83.3[/tex]
E cell :
[tex]\tt E_{cell}=2-\dfrac{0.0592}{3}log~83.3\\\\E_{cell}=1.962~V[/tex]
