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The human body has ~37 trillion cells ā more than the number of stars in the Milky Way.
In the electrochemical cell Zn ā£ | ā£ ZnSO 4 ( 0.1 M ) ā£ ā£ _4(0.1 M) || 4 ā ( 0.1 M ) ā£ā£ CuSO 4 ( 1.0 M ) ā£ _4 (1.0 M)| 4 ā ( 1.0 M ) ā£ Cu , , , the emf of this Daniel cell is E 1 . E_1. E 1 ā . When the concentration of ZnSO 4 _4 4 ā is changed to 1.0 M and that of CuSO 4 _4 4 ā changed to 0.01 M, the emf changed to E 2 . E_2. E 2 ā . From the following, which one is the relationship between E 1 E_1 E 1 ā and E 2 ( E_2 ( E 2 ā ( Given: R T F = 0.059 ) ? \frac{RT}{F} = 0.059)? F RT ā = 0.059 )?
medium
Electrochemistry
2017
chemistry
E 1 < E 2 E_1 < E_2 E 1 ā < E 2 ā
E 1 > E 2 E_1 > E_2 E 1 ā > E 2 ā
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 ā ā R T n F ln ā” Q \\
E = E^\circ - \frac{RT}{nF} \ln Q \\
E = E ā ā n F RT ā ln Q
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E 2 = 0 ā E 1 E_2 = 0 \neq E_1 E 2 ā = 0 ī = E 1 ā
E 1 = E 2 E_1 = E_2 E 1 ā = E 2 ā
Where:
ā¢
is the cell potential (emf).
ā¢
is the standard cell potential.
ā¢
is the universal gas constant.
ā¢
is the temperature in Kelvin.
ā¢
is the number of moles of electrons transferred in the reaction.
ā¢
is Faraday's constant.
ā¢
is the reaction quotient.
For the given cell reaction:
Zn
Cu
2 + ā ^{2+} \rightarrow 2 + ā Zn
Cu
The standard cell potential
is calculated as:
E ā = E c a t h o d e ā ā E a n o d e ā \newline E^\circ = E^\circ_{cathode} - E^\circ_{anode} \\
E ā = E c a t h o d e ā ā ā E an o d e ā ā Given that
R T F = 0.059 , \frac{RT}{F} = 0.059, F RT ā = 0.059 , we can write the Nernst equation as:
E = E ā ā 0.059 n log ā” Q \\
E = E^\circ - \frac{0.059}{n} \log Q \\
E = E ā ā n 0.059 ā log Q For the first condition:
ā¢ Concentration of ZnSO
M
ā¢ 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 ā ā 0.059 2 log ā” 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 ā = [ C u 2 + ] [ Z n 2 + ] ā = 1.0 0.1 ā = 0.1 E 1 ā = E ā ā 2 0.059 ā log 0.1 For the second condition:
ā¢ Concentration of ZnSO
M
ā¢ Concentration of CuSO
4 = 0.01 _4 = 0.01 4 ā = 0.01 M
The reaction quotient
is:
Q 2 = [ Z n 2 + ] [ C u 2 + ] = 1.0 0.01 = 100 E 2 = E ā ā 0.059 2 log ā” 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 ā = [ C u 2 + ] [ Z n 2 + ] ā = 0.01 1.0 ā = 100 E 2 ā = E ā ā 2 0.059 ā log 100 Now, calculate the difference between
and
E 2 : E 1 = E ā ā 0.059 2 log ā” 0.1 E 2 = E ā ā 0.059 2 log ā” 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 ā : E 1 ā = E ā ā 2 0.059 ā log 0.1 E 2 ā = E ā ā 2 0.059 ā log 100 Since
log ā” 0.1 = ā 1 \log 0.1 = -1 log 0.1 = ā 1 and
log ā” 100 = 2 , \log 100 = 2, log 100 = 2 , we have:
E 1 = E ā + 0.059 2 ā
1 E 2 = E ā ā 0.059 2 ā
2 E 1 = E ā + 0.0295 E 2 = E ā ā 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 ā = E ā + 2 0.059 ā ā
1 E 2 ā = E ā ā 2 0.059 ā ā
2 E 1 ā = E ā + 0.0295 E 2 ā = E ā ā 0.059 Comparing
and
E 2 : E 1 = E ā + 0.0295 E 2 = E ā ā 0.059 E_2: \\
E_1 = E^\circ + 0.0295 \\
E_2 = E^\circ - 0.059 \\
E 2 ā : E 1 ā = E ā + 0.0295 E 2 ā = E ā ā 0.059 Thus,
E 1 > E 2 . E_1 > E_2. \\
E 1 ā > E 2 ā . Therefore, the correct relationship is: Option 2:
E 1 > E 2 . E_1 > E_2. E 1 ā > E 2 ā .