Consider the cell pt | h2 (1 atm) | h + (? m) || hg2cl2(s) | cl− (1 m) | hg 2 h+ + 2 e − → h2 e 0 = 0. 00 v hg2cl2 + 2 e − → 2 hg + 2 cl− e 0 = 0. 268 v if the measured cell potential for the cell is 0. 416 volts, what is the ph of the solution?.

1. c

2. g

3. c

4. h

5. a

explanation:

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