Calculate the emf of the cell in which the following reaction takes place,
`Ni(s) + 2Ag^(+) ( 0.002M) rarr Ni^(2+) ( 0.160M) + 2Ag(s) `
Given that `E_("cell")^(@) = 1.05V`

Formulate the galvanic cell in which the following reaction takes place. Zn (s) + 2Ag^(+) ( aq) rarr Zn^(2+) (aq) + 2Ag(s) (i) Which one of its electrodes is negatively charged ? (ii) The reaction taking place at each of its electrode. (iii) The carriers of current within this cell.

Calculate the emf of the cell in which the following reaction takes place at two electrodes. ZnrarrZn^(2+)"_((aq))+2e^- , E^@=+0.76V Ag^+"_((aq))+e^(-)rarrAg , E^@=+0.80V

Calculate the standard cell potential of a galvanic cell in which the following reaction takes place 2Cr(s) + 3Cd^(2+) ( aq) rarr 2Cr^(3+) ( aq) + 3Cd(s) Calculate Delta_( r) G^(@) and equilibrium constant K of the above reaction at 25^(@)C . [Given, E_(Cr^(3+) //Cr)^(@) = - 0.74 V, E_(Cd^(2+)//Cd)^(@) = - 0.40V, 1F = 96500 C "mol"^(-1)]

Calculate the emf of the following cell at 25^(@) C Ag(s) |Ag^(+) ( 10^(-3) M ) | | Cu^(2+) ( 10^(-1) M ) | Cu (s) Given, E_("cell")^(@) = + 0.46 V and log 10^(n) = n .

In the button cells, widely used in watches and other devices, the following reaction takes place Zn (s) + Ag_(2) O(s) + H_(2) O (l) rarrZn^(2+) (aq) + 2Ag(s) + 2OH^(-) (aq) Determine E^(@) and Delta G^(@) for the reaction. (Given, E_(Ag^(+)//Ag)^(@) = + 0.80 V, E_(Zn^(2+)//Zn)^(@) = -0.76 V )

Calculate the standard emf of the cell having the cell reaction. Zn(s) + Co^(2+) ( aq) rarr Zn^(2+) ( aq) + Co(s) E_(Zn//Zn^(2+)) ^(@) = 0.76V, E_(Co//Co^(2+))^(@) = 0.25V

The cell in which the following reaction occurs, 2Fe^(3+) (aq) + 2I^(-)(aq) rarr 2Fe^(2+) (aq) + I_(2)(s) has E_("cell")^(@) = 0.236V at 298K , calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Calculate the cell emf and Delta_(r ) G^(@) for the cell reation at 25^(@)C . Zn(s) | Zn^(@+) ( 0.1 M )|| Cd^(2+) ( 0.01M) | Cd(s) Given, E_(Zn^(2+) //Zn)^(@) = - 0.763V , E_(Cd^(2+) //Cd) ^(@) = - 0.403 V 1F = 96500 C "mol"^(-1) R = 8.314 JK^(-1) "mol"^(-1) ], Find E_("cell")^(@) = E_("cathode")^(@) - E_("anode")^(@) then Delta _(r ) G^(@) by using formula, Delta _(r ) G^(@) = - n FE_("cell")^(@)