Four metals A, B, C, and D are placed into four beakers with 1.00 M solutions of
ID: 532933 • Letter: F
Question
Four metals A, B, C, and D are placed into four beakers with 1.00 M solutions of their positive ions A+, B2+, C3+, and D2+. The cells are connected with the following results:
(a) Write the four possible reduction half-reactions of the ions and arrange them from strongest to weakest as oxidizing agents. my answer = A+ > D2+ > C3+ > B2+
(b) Which metal is the weakest reducing agent? my answer = A+
(c) Write the balanced equation that would represent the reaction that occurs when A and B are connected.
my answer = 2A + B2+ --> 2A+ + B
(d) Predict the voltage when C and D are connected. my answer = I have no idea!!
Are the answers that I do have actually correct, or am I thinking about this all wrong? And how do I solve part (d)?? I've been strugging with this practice problem for days, if you could please show steps and explain so that I can learn for next time that would be much appreciated!
Anode Cathode Eocell A B 0.28 V A C 0.44 V D A 0.22 VExplanation / Answer
(a)
Strongest to weakest oxidizing agents:
C3+ > B2+ > A+ > D2+
(b)
Weakest reducing agent is the strongest oxidizing agent, which is C3+
(c)
2A(s) + B2+ ----> 2A+(aq) + B(s)
(d)
When C and D are connected, Ecell = 0.44 + 0.22 = 0.66 V
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