A) When the Cu2+ concentration is
1.07 M, the observed cell potential at 298K for an
electrochemical cell with the following reaction is
1.607V. What is the
Mn2+ concentration?
Cu2+(aq) +
Mn(s)-->
Cu(s) +
Mn2+(aq)
Answer: _____ M
B) When the Cu2+ concentration is
5.71×10-4 M, the observed cell
potential at 298K for an electrochemical cell with the following
reaction is 2.609V. What is the
Mg2+ concentration?
Cu2+(aq) +
Mg(s)--->
Cu(s) +
Mg2+(aq)
Answer: _____ M
C) When the Cu2+
concentration is 1.08 M, the observed cell
potential at 298K for an electrochemical cell with the following
reaction is 2.064V. What is the
Al3+ concentration?
3Cu2+(aq)
+
2Al(s)--->3Cu(s)
+
2Al3+(aq)
Answer: _____ M
a)
Cu2+(aq) + Mn(s)--> Cu(s) + Mn2+(aq)
E0cell = 1.517 V
Ecell = E0cell - 0.0591/nlog[Mn2+]/[cu2+]
1.607 = 1.517 -(0.0591/2)log(x/1.07)
[Mn2+] = 0.000963 M
b) Cu2+(aq) + Mg(s)---> Cu(s) + Mg2+(aq)
E0cell = 2.707 V
Ecell = E0cell - 0.0591/nlog[Mg2+]/[cu2+]
2.609 = 2.707 -(0.0591/2)log(x/(5.71*10^-4))
[Mg2+] = 1.183 M
C) 3Cu2+(aq) + 2Al(s)--->3Cu(s) + 2Al3+(aq)
E0cell = 1.997 V
Ecell = E0cell - 0.0591/nlog[Al3+]^2/[cu2+]^3
2.064 = 1.997 -(0.0591/6)log(x^2/1.08^3)
[Al3+] = 4.46*10^-4 M
Get Answers For Free
Most questions answered within 1 hours.