Question

4)Calculate the E_cell of: Al(s) + Cu²⁺(aq)(0.010M) → Al³⁺(aq)(1.0M) + Cu(s) (is this balanced?)

A. 2.06 B. 1.94 C. -2.06

5)Calculate the E^ocell for: Fe(s) + Br₂(g) →   Fe²⁺(aq) +Br^-(aq) ; is this electrolytic or voltaic?

A.-1.52V ; voltaic B.1.52V ; electrolytic C.1.52V ; voltaic D.-1.52V ; electrolytic

6)Calculate the E_cell of: Au(s) ; Au³⁺(aq) // Sn²⁺(aq) ; Sn(s)    is this electrolytic or voltaic?

A.1.64 ; electrolytic B.1.64 ; voltaic C.-1.64 ; voltaic D.-1.64 ; electrolytic

Answer 1

there are three questions. The data of electrode potential are not given with question so i am using symbols, calculation has to been done at your end using the electrode potential values.

4) its not balanced . the balanced equation is as follows :

2Al(s) +3 Cu²⁺(aq)(0.010M) → 2Al³⁺(aq)(1.0M) + 3Cu(s)

Ecell = E^o_Cell - 0.0591/6log(Al⁺³)²/ (Cu²⁺)³

        = E^o_Cell - 0.0591/6log(1)²/ (0.01)³

       = E^o_Cell - 0.0591x6/6 = E^o_Cell -0.0591 volt

      = (1.662 + 0.337 ) -0.0591 = 1.94 volt

5)       Fe(s) + Br₂(g) →   Fe²⁺(aq) +2Br^-(aq)

E^o_cell = (E^o Fe/Fe²⁺ )anode + (E^o Br₂/Br^-1 )cathode

           =        0.44                       +   1.066    = 1.5 voltaic

6) its a not avoltacic cell as its represented in wrong order.

the balanced reaction is as folows :

Sn should be anode and gold electrode should be cathode .

The E^o_cell   = (E^o_Au/Au⁺³ )anode + (E^o_Sn⁺²/Sn)cathode

               = 1.53 +   -0.13 = 1.4 volt