A solution contains Fe2+, Fe3+ and I$$-$$ ions. This solution was treated with iodine at 35oC. Eo for Fe3+/Fe2+ is + 0.77 V and Eo for I2/2I$$-$$ = 0.536 V.
The favourable redox reaction is
A
I2 will be reduced to I$$-$$
B
there will be no redox reaction
C
I$$-$$ will be oxidised to I2
D
Fe2+ will be oxidised to Fe3+
Explanation
Since the reduction potential of Fe3+/Fe2+
is greater than that of I2
/I–
, Fe3+ will be reduced
and I–
will be oxidised.
2Fe3+ + 2I– $$ \to $$ 2Fe2+ + I2
3
AIPMT 2011 Prelims
MCQ (Single Correct Answer)
The electrode potentials for Cu2+(aq) + e$$-$$ $$ \to $$ Cu+(aq) and Cu+(aq) + e$$-$$ $$ \to $$ Cu(s) are + 0.15 V and + 0.50 V respectively.
$$ \Rightarrow $$ Eo = $${{1 \times 0.15 + 1 \times 0.50} \over 2}$$ = 0.325 V
4
AIPMT 2011 Prelims
MCQ (Single Correct Answer)
Standard electrode potential for Sn4+/Sn2+ couple is + 0.15 V and that for the Cr3+/Cr couple is $$-$$ 0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be
A
+ 1.19 V
B
+ 0.89 V
C
+ 0.18 V
D
+ 1.83 V
Explanation
E°Sn4+/Sn2+ = 0.15 V
E°Cr3+/Cr = –0.74 V
E°cell = E°cathode – E°anode
= 0.15 – (– 0.74) = 0.15 + 0.74 = 0.89 V
Questions Asked from Electrochemistry
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