What is the equilibrium constant (K) at 25°C for the following cell reaction? Fe(s) + Cd2+(aq)------>Fe2+(aq)+ Cd(s). E0 cell = 0.010 V. Please show all steps clearly to solution. Thanks

For the galvanic (voltaic) cell Fe(s) + Mn²⁺(aq) ⟶ Fe²⁺(aq) + Mn(s) (E° = 0.77 V at 25 °C), what is [Fe²⁺] if [Mn²⁺] = 0.030 M and E = 0.78 V? Assume T is 298 K

Sn(s) | Sn2+ (aq, 1.8 M) || Agt(aq, 0.055 M) | Ag(s) Consider the cell notation of a galvanic cell given above and the following reduction potentials: Sn- (aq) + 2e- → Sn(s) E° = -0.14 V Ag (aq) + e- → Agis) ED=0.80 V 1. The number of electron transferred in the net reaction is [Select] 2. The E is [Select] 3. The value for Q is (Select] 4. The cell potential for the cell is [Select] V.

In a Voltaic cell composed of CuôCu(NO3)2 and AgôAgNO3 half-cells, what is the spontaneous cell reaction and the Eocell, respectively? The standard reduction potentials are as follows:                    Ag+(aq)  +  e–  -->  Ag(s);       Eo = 0.80 V                  Cu2+(aq)  + 2e–  -->  Cu(s);      Eo = 0.34 V          (A) 2Ag(s) + Cu2+(aq)  -->  2Ag+(aq) +  Cu(s);    Eocell = 1.14 V          (B) 2Ag(s) + Cu2+(aq)  -->  2Ag+(aq) +  Cu(s);    Eocell = 1.94 V          (C) Cu(s) + 2Ag+(aq)  -->  Cu2+(aq)  +  2Ag(s);  Eocell = 0.46 V          (D) Cu(s) + 2Ag+(aq)  -->  Cu2+(aq)  +  2Ag(s);  Eocell = 1.26 V