What will be the emf of the following cell at 25^{\circ} {\C} ? {\Fe} \/ {\Fe}^{2+}(0.001 {\M})|| {\H}^{+}(0.01 {\M}) ∣ {\H}_2( {\g}) (1 {\Bar}) ∣ {\Pt}(s) E_{ ( {\Fe}^{2+} \/ {\Fe}) }^{\circ}=-0.44 {\V} ; E_{ ( {\H}^{+} \/ {\H}_2) }^{\circ}=0.00 {\V}

COMEDK Exam 2022 Solved Paper

Question : 102 of 180

Marks: +1, -0

What will be the emf of the following cell at 25∘C ?
Fe∕Fe2+(0.001M)||
H+(0.01M)∣H2(g)
(1 Bar)∣Pt(s)
E(Fe2+∕Fe)∘=−0.44V;E(H+∕H2)∘=0.00V

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