Calculate the electrode potential given that E° `(Zn^(2+) // Zn)`=`-0.76V`, when a zinc rod is dipped in 0.1 M solution of `ZnSO_(4)`.The salt is 95% dissociated at this solution at 298K.

What does the negative sign in the expression E_(Zn^(2+) //Zn)^(@) = -0.76V mean ?

Calculate the electrode potential of a copper wire dipped in 0.1M CuSO_4 solution at 298 K. The standrd electrode potential of copper is +0.34 volt.

A monotropic acid in 1.00 M solution is 0.01% ionised. What is the dissociation constant of the acid ?

Formic acid is 4.6% dissociated in a 0.1 N solution at 20^@C . The ionisation constant of formic acid is :

The reduction electrode potential E of 0.1 M solution of M^+ ions (E^o(rp) = -2.36V ) is :

A strip of nickel metal is dipped in a 1 molar solution of Ni (NO_(3))_(2) and a strip of silver metal is dipped in a 1 molar solution of AgNO_(3) . An electrochemical cell is created when the two solutions are joined by salt bridge and two strips are joined by wire to a voltmeter. Answer the following questions. Calculate the cell potential (E_("cell")) at 25^(@)C for the cell if the initial concentration of Ni (NO_(3))_(2) , is 0.100 molar and the initial concentration of AgNO_(3) is 1.00 molar. [E_(Ni^(2+)//Ni)= -0.25V, E_("Ag^(+)//Ag)=0.80V, log 10^(-1)= -1]