The reduction potential for the following non-standard half cell
at 298K is______ volts.
2H+
(5.17×10-3M) +
2e- yields
H2
(1.08atm)
Sol .
As Reaction :
2H+ ( 5.17 × 10-3 M ) + 2e- <---> H2 ( 1.08 atm )
So, Reaction Quotient = Q
= p(H2) / [H+]2
= 1.08 / (5.17 × 10-3 )2 = 0.04040 × 106
Also , Standard Reduction potential for this half cell
= E° = 0
So ,
Reduction potential for this half cell
= E = E° - (0.0591 / n) × log(Q)
where n = no. of electrons transferred = 2
So ,
E = 0 - (0.0591/2) × log(0.04040 × 106 )
= - 0.136 V
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