Question

What concentration (in g/L) of Cu2+ is required to produce a half-cell potential of 0.32 V. The reference electrode is a SHE.

Answer 1

Voltmeter 6 Anode H₂ SHE Cathode Oxidation half-reaction reduction half-reaction Cu aq) + B é 21st 199) + 2€ 19/ When we close the circuit this time, the measured potential for the cell is negative, rather than positive. The negative value of E cell indicates that the direction of spontaneous electron flow is the positive Ecell, are the reduction of cet to cu at the copper electrode. The copper electrode gains mass at the reaction proceeds, and H₂ is oxidized to it at the platinum electrode. In this cell, the copper strip is the cathode and the hydrogen dectrode is the anode. The cell diagram therefore is written with the SHE on the left and cul cu couple on the right: I

P 18/H219, 1atm) / rit (aq, 1 M) 1 cut caq, IM) 1 cucco The half-cell reactions and potentials of to Spontaneous reaction are as follows: Cathode: concag trê Cung Ecathode =0.34v Anode: H₂ 1g, 24 ta tré Eauode to 1. Overall: H₂ g) + Cung) → 2H+ (aq) + Cusl® E cell = Ecathode - Eaude = 0.34v Thus the Standard electrode potential for the cat / cu couple is 0.34 V. from equation 0, to produce 0.34v, 1 mole of celt is required. 0.34 v I mole Cult 0.32 V ? = 0.32 x1 =0.94) mole - 0.34 required=0.941*66.5 = 597691 Molecular of cert willos do cu = 635 - 125 ..Amount a