When E cell is positive the redox reaction of the cell is spontaneous G will be

When e cell is positive the redox reaction of the

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When E ° cell is positive , the redox reaction of the cell is spontaneous ( G ° will be negative). When E ° cell is negative , the redox reaction of the cell is nonspontaneous ( G ° will be positive). Zn( s ) + Cu 2+ ( aq ) Zn 2+ ( aq ) + Cu( s ) spontaneous Cu( s ) + Zn 2+ ( aq ) Cu 2+ ( aq ) + Zn( s ) nonspontaneous Cd 2+ ( aq ) + 2e Cd ( s ) E ° = –0.40 V Cr 3+ ( aq ) + 3e Cr ( s ) E ° = –0.74 V Cd is the stronger oxidizer Cd will oxidize Cr 2e + Cd 2+ (1 M ) Cd (s) Cr (s) Cr 3+ (1 M ) + 3e Anode (oxidation): Cathode (reduction): 2Cr ( s ) + 3Cd 2+ (1 M ) 3Cd ( s ) + 2Cr 3+ (1 M ) × 2 × 3 Example: What is the standard cell potential of a galvanic (spontaneous) cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution at 25 °C? E ° cell = E ° cathode E ° anode E ° cell = –0.40 V ( 0.74 V) E ° cell = +0.34 V The half-reaction higher in the table (larger value of ) will occur as written (reduction), while the other will be reversed.
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12 Predicting whether a Metal Will Dissolve in an Acid • Most metals dissolve in acids – if the reduction of the metal ion is easier than the reduction of H + ( aq ); – if their ion reduction reaction lies below H + reduction on the table. • Almost all metals will dissolve in HNO 3 . – Having N reduced rather than H. – Au and Pt dissolve in HNO 3 + HCl. Zn( s ) + 2H + ( aq ) Zn 2+ ( aq ) + H 2 ( g ) Spontaneity of Redox Reactions G = – nFE cell G ° = – nFE ° cell n = number of moles of electrons in the reaction F = 96,485 J V mol = 96,485 C/mol G 0 = – RT ln K = nFE ° cell E ° cell = RT nF ln K (8.314 J/K mol)(298 K) n (96,485 J/V mol) ln K = = 0.0592 V n log K E ° cell
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13 Spontaneity of Redox Reactions For a spontaneous reaction, one that proceeds in the forward direction with the chemicals in their standard states: G ° < 0 E ° > 0 K > 1 Fe 2+ + 2e Fe 2Ag 2Ag + + 2e Oxidation (anode): Reduction (cathode): n = 2 K = 1.28 × 10 –42 Example: Calculate the equilibrium constant and for the following reaction at 25 °C. Fe 2+ ( aq ) + Ag( s ) Fe( s ) + 2Ag + ( aq ) = 0.0592 V n log K E ° cell G 0 = – RT ln K = nFE ° cell E ° cell = E ° cathode E ° anode E ° cell = –0.44 V (0.80 V) E ° cell = –1.24 V 42 9 . 41 10 28 . 1 10 9 . 41 V 0592 . 0 V) 24 . 1 ( 2 V 0592 . 0 log log V 0592 . 0 × = = = = = = k nE K K n E o cell o cell G 0 = – nFE ° cell G 0 = –(2)(96500 C/mol)(–1.24 V) G 0 = 239000 J/mol = 239 kJ/mol E ° cell = –1.24 V G 0 = 239 kJ/mol Reaction is nonspontaneous
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14 The Effect of Concentration on Cell Potential G = G ° + RT ln Q G = – nFE G ° = – nFE° nFE = – nFE ° + RT ln Q Nernst equation At 298 K 0.0592 V n log Q E = Used for non-standard conditions RT nF ln Q E = Example: Will the following reaction occur spontaneously at 25 ° C when [Fe 2+ ] = 0.60 M and [Cd 2+ ] = 0.010 M ? Fe 2+ ( aq ) + Cd ( s ) Fe ( s ) + Cd 2+ ( aq ) Fe 2+ + 2e Fe Cd Cd 2+ + 2e Oxidation (anode): Reduction (cathode): n = 2 E ° cell = E ° cathode E ° anode E ° cell = –0.44 V – (–0.40 V) E ° cell = –0.04 V The reaction is nonspontaneous at standard conditions [ ] [ ] V 01 . 0
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