Consider the galvanic cell described by (N and M a...

Question

Consider the galvanic cell described by (N and M are metals):

N(s) | N²⁺(aq) || M⁺(aq) | M(s)

If E°_cathode = 0.182 V and E°_anode = 1.411 V, and [N²⁺(aq)] = 0.931 M and [M⁺(aq)] = 0.655 M, what is E_cell, using the Nernst equation? ____ V

Jules Bruno · Accepted Answer

We're given the following galvanic cell:N(s) | N2+(aq) || M+(aq) | M(s)When writing a cell notation, we use the following format – “as easy as ABC”Lose               GainElectron         ElectronsOxidation       Reduction• lose electrons → oxidation → anode• gain electrons → reduction → cathode[readmore]N(s) | N2+(aq) || M+(aq) | M(s)      (anode)                               (cathode)Step 1: Write the overall reaction.• Anode ▪ oxidation → lose electrons▪ lose electrons → from more negative to less negative (more positive) oxidation stateN(s) → natural state → oxidation state = 0N2+(aq) → ion → oxidation state = charge → oxidation state = +2Anode Reaction:        N(s) → N2+(aq)                                                        (product)• Cathode ▪ reduction → gain electrons▪ gain electrons → from less negative to more negative (less positive) oxidation stateM(s) → natural state → oxidation state = 0M+(aq) → ion → oxidation state = charge → oxidation state = +1Cathode Reaction:       M+(aq) → M(s)                                            (reactant)               N(s) → N2+(aq) + 2 e-      [M+(aq) + e- → M(s)] x 2                N(s) → N2+(aq) + 2 e- 2 M+(aq) + 2 e- → 2 M(s)_____________________ N(s) + 2 M+(aq) → N+(aq) + 2 M(s) n = 2 e-Step 2: Determine the standard cell potential.E°cell = E°cathode – E°anodeE°cathode = 0.182 V and E°anode = 1.411 VE°cell = 0.182 V – 1.411 VE°cell = –1.229 VStep 3: Calculate Ecell, using the Nernst equation.Ecell=E°cell-(0.05916 Vn) log QEcell=E°cell-(0.05916 Vn) log (productsreactants)Ecell=E°cell-(0.05916 Vn) log ([N2+][M+]2)Ecell=-1.229 V-(0.05916 V2) log ([0.931][0.655]2)Ecell=-1.229 V-(0.05916 V2) log (0.9310.429025)Ecell=-1.229 V-(0.02958 V) log (2.170036711)Ecell=-1.229 V-(0.02958 V) (0.33647)Ecell=-1.229 V-9.9527×10-3 VEcell = –1.239 VThe Ecell of the given galvanic cell is  –1.239 V.

Problem: Consider the galvanic cell described by (N and M are metals):N(s) | N2+(aq) || M+(aq) | M(s)If E°cathode = 0.182 V and E°anode = 1.411 V, and [N2+(aq)] = 0.931 M and [M+(aq)] = 0.655 M, what is Ecell, using the Nernst equation? ____ V

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Problem: Consider the galvanic cell described by (N and M are metals):N(s) | N2+(aq) || M+(aq) | M(s)If E°cathode = 0.182 V and E°anode = 1.411 V, and [N2+(aq)] = 0.931 M and [M+(aq)] = 0.655 M, what is Ecell, using the Nernst equation? ____ V

FREE Expert Solution

Problem Details

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