The value of Eºcell for the reaction Zn(s) + Cu 2+ (aq) <=> Zn 2+ (aq) + Cu(s) i
ID: 523405 • Letter: T
Question
The value of Eºcell for the reaction Zn(s) + Cu2+(aq) <=> Zn2+(aq) + Cu(s) is 1.10 V. What is the value of E for a cell in which [Cu2+] = 1.0 x 10-3 M and [Zn2+] = 1.0 M? Answer: 1.01 V Please explain step by step in a clear way! The value of Eºcell for the reaction Zn(s) + Cu2+(aq) <=> Zn2+(aq) + Cu(s) is 1.10 V. What is the value of E for a cell in which [Cu2+] = 1.0 x 10-3 M and [Zn2+] = 1.0 M? Answer: 1.01 V Please explain step by step in a clear way! The value of Eºcell for the reaction Zn(s) + Cu2+(aq) <=> Zn2+(aq) + Cu(s) is 1.10 V. What is the value of E for a cell in which [Cu2+] = 1.0 x 10-3 M and [Zn2+] = 1.0 M? Answer: 1.01 V Please explain step by step in a clear way!Explanation / Answer
Number of electron being transferred in balanced reaction is 2
So, n = 2
Use:
E = Eo - (2.303*RT/nF) log {[Zn2+]^1/[Cu2+]^1}
2.303*R*T/n = 2.303*8.314*298.0/F= 0.0591
E = Eo - (0.0591/n) log {[Zn2+]^1/[Cu2+]^1}
E = 1.1 - (0.0591/2) log (1.0^1/0.001^1)
E = 1.1-(0.089)
E = 1.011 V
Answer: 1.011 V
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.