Standard cell potential and spontaneity

Calculate E°cell = E°(cathode) - E°(anode) from standard electrode potentials, identify which half-cell is oxidized, and predict whether a redox reaction is spontaneous in the forward or reverse direction from the sign of E°cell.

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Zadanie 37Paper 1A, maj 2025, TZ2
wzorowane1 pktzamknięteśrednie (szac.)

What is the standard cell potential, EcellE^\ominus_\text{cell}, of the following cell? PbX2+(aq)+2eXPb(s)\ce{Pb^2+(aq) + 2e- <=> Pb(s)} E=0.13 VE^\ominus = -0.13\ \mathrm{V} ZnX2+(aq)+2eXZn(s)\ce{Zn^2+(aq) + 2e- <=> Zn(s)} E=0.76 VE^\ominus = -0.76\ \mathrm{V} A. 0.89 V-0.89\ \mathrm{V} B. 0.63 V-0.63\ \mathrm{V}

Zadanie 39Paper 1A, maj 2025, TZ3
wzorowane1 pktzamknięteśrednie (szac.)

The table shows standard electrode potentials for three metals. Half-equation EE^\ominus / V MnX2+(aq)+2eXMn(s)\ce{Mn^2+(aq) + 2e- <=> Mn(s)} −1.18 CrX3+(aq)+3eXCr(s)\ce{Cr^3+(aq) + 3e- <=> Cr(s)} −0.74 CdX2+(aq)+2eXCd(s)\ce{Cd^2+(aq) + 2e- <=> Cd(s)} −0.40 Which combination undergoes a spontaneous…

Zadanie 35Paper 1A, listopad 2025, TZ3
wzorowane1 pktzamkniętełatwe (szac.)

A voltaic cell is made by connecting a manganese half-cell (Mn(s)\ce{Mn(s)} in MnX2+(aq)\ce{Mn^2+(aq)}) and a tin half-cell (Sn(s)\ce{Sn(s)} in SnX2+(aq)\ce{Sn^2+(aq)}) under standard conditions. MnX2+(aq)+2eXMn(s)\ce{Mn^2+(aq) + 2e- <=> Mn(s)} E=1.18 VE^\ominus = -1.18\ \mathrm{V}

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