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In the electrochemical cell Zn โฃ | โฃ ZnSO 4 ( 0.1 M ) โฃ โฃ _4(0.1 M) || 4 โ ( 0.1 M ) โฃโฃ CuSO 4 ( 1.0 M ) โฃ _4 (1.0 M)| 4 โ ( 1.0 M ) โฃ Cu , , , the emf of this Daniel cell is E 1 . E_1. E 1 โ . When the concentration of ZnSO 4 _4 4 โ is changed to 1.0 M and that of CuSO 4 _4 4 โ changed to 0.01 M, the emf changed to E 2 . E_2. E 2 โ . From the following, which one is the relationship between E 1 E_1 E 1 โ and E 2 ( E_2 ( E 2 โ ( Given: R T F = 0.059 ) ? \frac{RT}{F} = 0.059)? F RT โ = 0.059 )?
medium
Electrochemistry
2017
chemistry
E 1 < E 2 E_1 < E_2 E 1 โ < E 2 โ
E 1 > E 2 E_1 > E_2 E 1 โ > E 2 โ
Explanation To solve this problem, we will use the Nernst equation to calculate the emf of the cell under different conditions. \\
The Nernst equation for a cell is given by: E = E โ โ R T n F ln โก Q \\
E = E^\circ - \frac{RT}{nF} \ln Q \\
E = E โ โ n F RT โ ln Q
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E 2 = 0 โ E 1 E_2 = 0 \neq E_1 E 2 โ = 0 ๎ = E 1 โ
E 1 = E 2 E_1 = E_2 E 1 โ = E 2 โ
Where:
โข
is the cell potential (emf).
โข
is the standard cell potential.
โข
is the universal gas constant.
โข
is the temperature in Kelvin.
โข
is the number of moles of electrons transferred in the reaction.
โข
is Faraday's constant.
โข
is the reaction quotient.
For the given cell reaction:
Zn
Cu
2 + โ ^{2+} \rightarrow 2 + โ Zn
Cu
The standard cell potential
is calculated as:
E โ = E c a t h o d e โ โ E a n o d e โ \newline E^\circ = E^\circ_{cathode} - E^\circ_{anode} \\
E โ = E c a t h o d e โ โ โ E an o d e โ โ Given that
R T F = 0.059 , \frac{RT}{F} = 0.059, F RT โ = 0.059 , we can write the Nernst equation as:
E = E โ โ 0.059 n log โก Q \\
E = E^\circ - \frac{0.059}{n} \log Q \\
E = E โ โ n 0.059 โ log Q For the first condition:
โข Concentration of ZnSO
M
โข Concentration of CuSO
M
The reaction quotient
is:
Q 1 = [ Z n 2 + ] [ C u 2 + ] = 0.1 1.0 = 0.1 E 1 = E โ โ 0.059 2 log โก 0.1 \\
Q_1 = \frac{[{Zn}^{2+}]}{[{Cu}^{2+}]} = \frac{0.1}{1.0} = 0.1 \\
E_1 = E^\circ - \frac{0.059}{2} \log 0.1 \\
Q 1 โ = [ C u 2 + ] [ Z n 2 + ] โ = 1.0 0.1 โ = 0.1 E 1 โ = E โ โ 2 0.059 โ log 0.1 For the second condition:
โข Concentration of ZnSO
M
โข Concentration of CuSO
4 = 0.01 _4 = 0.01 4 โ = 0.01 M
The reaction quotient
is:
Q 2 = [ Z n 2 + ] [ C u 2 + ] = 1.0 0.01 = 100 E 2 = E โ โ 0.059 2 log โก 100 \\
Q_2 = \frac{[{Zn}^{2+}]}{[{Cu}^{2+}]} = \frac{1.0}{0.01} = 100 \\
E_2 = E^\circ - \frac{0.059}{2} \log 100 \\
Q 2 โ = [ C u 2 + ] [ Z n 2 + ] โ = 0.01 1.0 โ = 100 E 2 โ = E โ โ 2 0.059 โ log 100 Now, calculate the difference between
and
E 2 : E 1 = E โ โ 0.059 2 log โก 0.1 E 2 = E โ โ 0.059 2 log โก 100 E_2: \\
E_1 = E^\circ - \frac{0.059}{2} \log 0.1 \\
E_2 = E^\circ - \frac{0.059}{2} \log 100 \\
E 2 โ : E 1 โ = E โ โ 2 0.059 โ log 0.1 E 2 โ = E โ โ 2 0.059 โ log 100 Since
log โก 0.1 = โ 1 \log 0.1 = -1 log 0.1 = โ 1 and
log โก 100 = 2 , \log 100 = 2, log 100 = 2 , we have:
E 1 = E โ + 0.059 2 โ
1 E 2 = E โ โ 0.059 2 โ
2 E 1 = E โ + 0.0295 E 2 = E โ โ 0.059 \\
E_1 = E^\circ + \frac{0.059}{2} \cdot 1 \\
E_2 = E^\circ - \frac{0.059}{2} \cdot 2 \\
E_1 = E^\circ + 0.0295 \\
E_2 = E^\circ - 0.059 \\
E 1 โ = E โ + 2 0.059 โ โ
1 E 2 โ = E โ โ 2 0.059 โ โ
2 E 1 โ = E โ + 0.0295 E 2 โ = E โ โ 0.059 Comparing
and
E 2 : E 1 = E โ + 0.0295 E 2 = E โ โ 0.059 E_2: \\
E_1 = E^\circ + 0.0295 \\
E_2 = E^\circ - 0.059 \\
E 2 โ : E 1 โ = E โ + 0.0295 E 2 โ = E โ โ 0.059 Thus,
E 1 > E 2 . E_1 > E_2. \\
E 1 โ > E 2 โ . Therefore, the correct relationship is: Option 2:
E 1 > E 2 . E_1 > E_2. E 1 โ > E 2 โ .