In a certain experiment, a Galvanic cell has a cell emf = 0.54 V at 25 °C. Given that [Zn2+] = 1.0 M and PH2 = 1.0 atm. Calculate [H+], AND PH. Zn (s) | Zn2+ (1.0 M) || H+ (?? M) | H₂ (1.0 atm) | Pt (s) Oxidation: Reduction: (2e)+2H+ H2 E° = E₁₁+/H2 - Ezn/ZR+ 2+ E° = 0 (-0.76) Zn Zn2+ +(2e- n = 2 E = E°- 0.0257 V In Q n 0.54 = 0.76 0.0257 [Zn2+]PH2 In 2 [H+]2 E° = +0.76 V 0.0257 (1.0)(1.0) -0.22 = 1 In 2 (x)² 17.1 = In (1.0) (x)² e17.1 = (1.0) (x)² x = [H+] = 1.935 x 10-4 M NOTE: use "log" for pH calcs! pH = -log(1.935 x 10 4 M) pH = 3.713
In a certain experiment, a Galvanic cell has a cell emf = 0.54 V at 25 °C. Given that [Zn2+] = 1.0 M and PH2 = 1.0 atm. Calculate [H+], AND PH. Zn (s) | Zn2+ (1.0 M) || H+ (?? M) | H₂ (1.0 atm) | Pt (s) Oxidation: Reduction: (2e)+2H+ H2 E° = E₁₁+/H2 - Ezn/ZR+ 2+ E° = 0 (-0.76) Zn Zn2+ +(2e- n = 2 E = E°- 0.0257 V In Q n 0.54 = 0.76 0.0257 [Zn2+]PH2 In 2 [H+]2 E° = +0.76 V 0.0257 (1.0)(1.0) -0.22 = 1 In 2 (x)² 17.1 = In (1.0) (x)² e17.1 = (1.0) (x)² x = [H+] = 1.935 x 10-4 M NOTE: use "log" for pH calcs! pH = -log(1.935 x 10 4 M) pH = 3.713
Chapter18: Introduction To Electrochemistry
Section: Chapter Questions
Problem 18.18QAP
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![In a certain experiment, a Galvanic cell has a cell emf = 0.54 V at 25 °C.
Given that [Zn2+] = 1.0 M and PH2 = 1.0 atm. Calculate [H+], AND PH.
Zn (s) | Zn2+ (1.0 M) || H+ (?? M) | H₂ (1.0 atm) | Pt (s)
Oxidation:
Reduction:
(2e)+2H+ H2
E° = E₁₁+/H2 - Ezn/ZR+
2+
E° = 0 (-0.76)
Zn Zn2+ +(2e-
n = 2
E = E°-
0.0257 V
In Q
n
0.54 = 0.76
0.0257 [Zn2+]PH2
In
2
[H+]2
E° = +0.76 V
0.0257
(1.0)(1.0)
-0.22 =
1
In
2
(x)²
17.1 = In (1.0)
(x)²
e17.1 = (1.0)
(x)²
x = [H+] = 1.935 x 10-4 M
NOTE: use "log" for pH calcs!
pH = -log(1.935 x 10 4 M)
pH = 3.713](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F637a8243-98d2-45ca-8dd1-4e9d7d0f97ca%2F1dc147e3-f785-4d26-9bdc-aa1d87c8db5c%2Fg71qy2e.jpeg&w=3840&q=75)
Transcribed Image Text:In a certain experiment, a Galvanic cell has a cell emf = 0.54 V at 25 °C.
Given that [Zn2+] = 1.0 M and PH2 = 1.0 atm. Calculate [H+], AND PH.
Zn (s) | Zn2+ (1.0 M) || H+ (?? M) | H₂ (1.0 atm) | Pt (s)
Oxidation:
Reduction:
(2e)+2H+ H2
E° = E₁₁+/H2 - Ezn/ZR+
2+
E° = 0 (-0.76)
Zn Zn2+ +(2e-
n = 2
E = E°-
0.0257 V
In Q
n
0.54 = 0.76
0.0257 [Zn2+]PH2
In
2
[H+]2
E° = +0.76 V
0.0257
(1.0)(1.0)
-0.22 =
1
In
2
(x)²
17.1 = In (1.0)
(x)²
e17.1 = (1.0)
(x)²
x = [H+] = 1.935 x 10-4 M
NOTE: use "log" for pH calcs!
pH = -log(1.935 x 10 4 M)
pH = 3.713
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