From the following information, calculate the solubility product of `AgBr.`
`AgBr(s)+e^(-) rarr Ag(s) +Br^(c-)(aq), " "E^(c-)=0.07V`
`Ag^(o+)(aq)+e^(-) rarr Ag(s)," "E^(c-)=0.080V`

From the following information, calculate the solubility product of silver bromide ? {:(AgBr(s)+e^(-)hArr Ag(s)+Br^(-)(aq.)E^(@)=0.07),(Ag^(+)(aq.)+e^(-)hArr Ag(s)E^(@)=0.80 V):}

In a simple electrochemical cell, the half cell reduction reaction with their standard electrode potentials are : Pb(s)-2e^(-) rarr Pb^(2+)(aq) , E^(@)= - 0.30 V Ag(s) -e^(-) rarr Ag^(+) (aq), E^(@)= + 0.80 V Which of the following reaction takes place ?

Ag(s) |Ag^+ (aq) (0.01M) || Ag^+ (aq) (0.1M)| Ag(s) E^@ Ag(s) // Ag(aq) = 0.80 volt.

Calculate the equilibrium constant for the reaction Cu(s)+2Ag+(aq)rarrCu^(+2)(aq)+2Ag(s),E_("cell")^(@)=0.46V .

The cell used in watch has the following cell reaction. Calculate E^@ & DeltaG^@ for it. Zn(s) + Ag_2O(s) + H_2O (l) rarr 2Zn^(2+) (aq) + 2Ag (s) + 2OH^- (aq) Given E^@_(Ag^+//Ag) = +0.80 V , E^@_(Zn^(2+)//Zn) = -0.76 V

Calculate the equilibrium constant of the reaction : Cu(s)+2Ag^(+)(aq)rarr Cu^(2+)(aq)+2Ag(s): E_("cell")^(@)=0.46V

Consider half reactions with standard potentials 'i. Ag^(o+) (aq) + e^(o-)rarr Ag (s) E^0 = 0.8V , ii. I_2(s) + 2e^(o-)rarr 2I^(o-) (aq) E^0 = 0.53V , iii. Pb^(2o+) (aq) + 2e^(o-) rarr Pb (s) E^0 = -0.13V , iv. Fe^(2o+) (aq) + 2e^(o-) rarr Fe (s) E^0 = -0.44V The strongest oxidising and reducing agents respectively are

The standard reduction potentials for two reactions are given below: AgCl(s) + e^(-) to Ag(s) + Cl^(-) , E^@ = 0.22 V Ag^(+)(aq) e^(-) +e^(-) to Ag(s) , E^@ = 0.80V The solubility product of AgCl under standard conditions of temperature (298 K) is given by