In another cell, the potential of the copper metal, copper(II) ion electrode was

Question

In another cell, the potential of the copper metal, copper(II) ion electrode was found to be 0.442 V relative to a silver metal, silver ion electrode, with the copper electrode being negative. If the potential of the silver, silver ion electrode, E_Ag^+, Ag^compositefunction is taken to be 0.000 V in oxidation or reduction, what is the value of the potential for the oxidation reaction, ?E_cell^compositefunction = E_oxid^compositefunction + E_red^compositefunction 0.442 volts If E_Ag^+, Ag red^compositefunction equals +0.799V, as in standard tables of electrode potentials, what is the value of the potential of the oxidation reaction of copper, ? __________volts The student adds 6 M NH_3, to the CuSO_4 solution in this second cell until the Cu^2+ ion is essentially all converted to Cu(NH_3)_3^2+ ion. The voltage of the cell, E_cell, goes up to 0.917Vand the Cu electrode is still negative. Find the residual concentration of Cu^2+ ion in the cell. (Use Eq. 4.) __________ M

Explanation / Answer

In this question giving copper electrode is negative that means copper is anode and another one silver electrode is cathode the cell constraction was Cu/Cu+2//Ag+/Ag

now formula for Ecell=Ecathode + Eanode named it as formula 1

For first question we need to find ECu/Cu+2 that means we need to find  Eanode and given that Ecell=0.442 V , EAg+/Ag = Ecathode= 0

then apply the values in formula 1 we get 0.442=0+Eanode there four Eanode=0.442 that means ECu/Cu+2 =0.442V

in second question They given EAg+/Ag = Ecathode=0.799 V then substitute the value in formula 1 we get

0.442=0.799+Eanode

there fore Eanode=-0.799+0.442

Eanode=-0.357V

ECu/Cu+2 =-0.357V

For third question answer we need to know cell equation is Cu+ 2Ag+ = Cu+2 + 2Ag

nernest formula E= Eo - 0.059/n log (p/r): p=product, r=reactant

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