A) When the Cu 2+ concentration is 1.07 M, the observed cell potential at 298K f

Question

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

Explanation / Answer

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

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