Determine the Eell of a voltaic cell composed of Cu(s) in a 1.00 M solution of Cu²+, and Cr(s) in a 1.00 M solution of Cr³+ at 25.00°C. Eº = 0.342 V Eº = -0.740 V Cu(s)/Cu²+ Cr(s)/Cr³+ V

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Determine the cell of a voltaic cell composed of Cu(s) in a 1.00 M solution of Cu²+, and Cr(s) in a
1.00 M solution of Cr³+ at 25.00°C.
Cu(s)/Cu²+
Eº = 0.342 V
Cr(s)/Cr³+
Eº = -0.740 V
V
Transcribed Image Text:Determine the cell of a voltaic cell composed of Cu(s) in a 1.00 M solution of Cu²+, and Cr(s) in a 1.00 M solution of Cr³+ at 25.00°C. Cu(s)/Cu²+ Eº = 0.342 V Cr(s)/Cr³+ Eº = -0.740 V V
Expert Solution
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We know that higher the standard reduction potential value, greater will be tendency to get reduce at cathode and lower the standard reduction potential value greater will be tendency to get oxidise at anode.

Therefore, from the given standard reduction potential of the copper electrode and chromium electrode. 

     E° of copper is than the E° of chromium 

Hence, copper will reduce and chromium will oxidise. 

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