could you help with this please. 3. Let\'s say that we didn\'t have access to a

Question

could you help with this please.

3. Let's say that we didn't have access to a standard H'/H, electrode. Instead, we made our half-cell potential measurements relative to the tin half-cell. (a) What would the Eon be for the half-reaction: Sn(s) Sn2+ (aq) + 2e ? Explain. (b) What would the E be for the oxidation half-reactions of (1) Cuis).(2) Ni(s), and (3) Ags)? (c) What would the Ecell be for the Cu-Ag electrochemical cell relative to the tin electrode? Provide both half-cell reactions, potentials relative to tin, and the net cell reaction. (d) We found a sandard IH /H, electrode and redid the calculations necessary for part c of this question. What would E'cell be now? (e) Compare the answers you got for parts c and d. What can you conclude about 207

Explanation / Answer

a)

We need first:

from reduction potentilas we get:

Sn2+ + 2 e Sn(s) 0.13

but we need the inverted

so when we invert this

the potential becomes positiv

Enew = --0.13 = 0.13 V

b)

E°ox for:

Ag+ + e Ag(s) +0.7996 --> inverted (xoidation) potential = -0.7996

Cu2+ + 2 e Cu(s) +0.337 --> inverted (xoidation) potential = -0.337

Ni2+ + 2 e Ni(s) 0.25 --> inverted (xoidation) potential = --0.25 = 0.25 V

C)

E°cell (Ag) = 0.7996 + 0.13 = 0.9296 V for Ag+ given Sn as reference

E°cell (cu) = 0.337 + 0.13 = 0.467 V for Cu given Sn as reference

d)

E°cell will be the same, since reference will be cancelled

e)

c and d --> are the SAME, since reference will be cancelled, the overal "difference" lies between Cu and Ag potential difference

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