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Question
Calculate emf of the cell at 25°C.
Zn(s) | Zn2+ (0.08 M) || Cr3+ (0.1 M) | Cr(s)
`E_(Zn)^0` = −0.76 V, `E_(Cr)^0` = −0.74 V
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Solution
Given: `E_(Zn)^0` = −0.76 V, `E_(Cr)^0` = −0.74 V
To find: Emf of the cell `(E_(cell))`
Formulae:
1) `E_(cell)^0 = E_(cathode)^0 - E_(anode)^0`
2) `E_(cell) = E_(cell)^0 - (0.0592 V)/n log_10 [["Product"]]/[["Reactant"]]`
Calculation:
\[\ce{[Zn_{(s)} -> Zn^{2+}_{(0.08 M)} + 2e-] × 3}\] (oxidation at anode)
\[\ce{[Cr^{3+}_{(0.1 M)} + 3e- -> Cr_{(s)}] × 2}\] (reduction at cathode)
___________________________________________________
\[\ce{3Zn_{(s)} + 2Cr^{3+}_{(0.1 M)} -> 3Zn^{2+}_{(0.08 M)} + Cr_{(s)}}\] (overall reaction)
Using formula (1),
`E_(cell)^0 = E_(cathode)^0 - E_(anode)^0`
`E_(cell) = E_(Cr)^0 - E_(Zn)^0`
= −0.74 V − (−0.76 V)
= 0.02 V
Using formula (2),
The cell potential is given by
`E_(cell) = E_(cell)^0 - (0.0592 V)/n log_10 [["Product"]]/[["Reactant"]]`
= `0.02 - (0.0592 V)/6 log_10 (0.08)^3/(0.1)^2`
= `0.02 - 9.867 × 10^-3 log_10 (5.12 × 10^-4)/(1 × 10^-2)`
= `0.02 - 9.867 × 10^-3 × bar 2. 7093`
= 0.02 − 9.867 × 10−3 × (−1.2907)
= 0.02 + 0.01273
= 0.03273 V
The emf of the cell is 0.03273 V.
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