Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG for the following redox reaction.Be sure your answer has the correct number of significant digits.2+SZn (s)+2103 (aq)+12H (aq)- 5Zn" (aq) +12 (s)+6H20 (I)to? Ag (aq)e- Ag (s)Al3+ (aq) 3e Al (s)0.7996-1.676Au (aq) e" Au (s)1.692Au3+ (aq) 3eAu (s)1.498Ba2+ (aq)2eBa (s)-2.912Br2 ()2e - 2Br (aq)1.066Ca2+ (aq)2e -- Са (s)-2.868Cl2 (g)2e » 2Cl (aq)1.35827Co2+ (aq) 2eCo (s)-0.28Co3+ (aq) e- Co2+ (aq)1.92Cr2+ (aq) 2eCr (s)-0.913Cr3+ (ag) 3e Cr (s)-0.744Cr3+ (aq) e- Cr2+ (aq)-0.407CrO42- (aq) 4H20 (I) +3eCr(OH)3 (s)50H- (aq)-0.13Cu2+ (aq)2e-Cu (s)0.3419Cu2+ (aq) e Cu (aq)0.153Cu (aq)e-Cu (s)0.521F2 (g)2e 2F (aq)Fe2+ (aq)2e2.866Fe (s)-0.447Fe3+ (aq)e"- Fe2+ (aq)0.771Fe3+ (aq) 3eFe (s)-0.0372H+ (aq) 2eH2 (g)0.0002H20 ()2e- H2 (g) + 20H (aq)-0.8277

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Asked Nov 16, 2019
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Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG for the following redox reaction.
Be sure your answer has the correct number of significant digits.
2+
SZn (s)+2103 (aq)+12H (aq)- 5Zn" (aq) +12 (s)+6H20 (I)
to
?
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Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG for the following redox reaction. Be sure your answer has the correct number of significant digits. 2+ SZn (s)+2103 (aq)+12H (aq)- 5Zn" (aq) +12 (s)+6H20 (I) to ?

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Ag (aq)e- Ag (s)
Al3+ (aq) 3e Al (s)
0.7996
-1.676
Au (aq) e" Au (s)
1.692
Au3+ (aq) 3e
Au (s)
1.498
Ba2+ (aq)2e
Ba (s)
-2.912
Br2 ()2e - 2Br (aq)
1.066
Ca2+ (aq)2e -
- Са (s)
-2.868
Cl2 (g)2e » 2Cl (aq)
1.35827
Co2+ (aq) 2e
Co (s)
-0.28
Co3+ (aq) e- Co2+ (aq)
1.92
Cr2+ (aq) 2e
Cr (s)
-0.913
Cr3+ (ag) 3e Cr (s)
-0.744
Cr3+ (aq) e- Cr2+ (aq)
-0.407
CrO42- (aq) 4H20 (I) +3e
Cr(OH)3 (s)
50H- (aq)
-0.13
Cu2+ (aq)2e-
Cu (s)
0.3419
Cu2+ (aq) e Cu (aq)
0.153
Cu (aq)e-Cu (s)
0.521
F2 (g)2e 2F (aq)
Fe2+ (aq)2e
2.866
Fe (s)
-0.447
Fe3+ (aq)e"- Fe2+ (aq)
0.771
Fe3+ (aq) 3e
Fe (s)
-0.037
2H+ (aq) 2e
H2 (g)
0.000
2H20 ()2e- H2 (g) + 20H (aq)
-0.8277
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Ag (aq)e- Ag (s) Al3+ (aq) 3e Al (s) 0.7996 -1.676 Au (aq) e" Au (s) 1.692 Au3+ (aq) 3e Au (s) 1.498 Ba2+ (aq)2e Ba (s) -2.912 Br2 ()2e - 2Br (aq) 1.066 Ca2+ (aq)2e - - Са (s) -2.868 Cl2 (g)2e » 2Cl (aq) 1.35827 Co2+ (aq) 2e Co (s) -0.28 Co3+ (aq) e- Co2+ (aq) 1.92 Cr2+ (aq) 2e Cr (s) -0.913 Cr3+ (ag) 3e Cr (s) -0.744 Cr3+ (aq) e- Cr2+ (aq) -0.407 CrO42- (aq) 4H20 (I) +3e Cr(OH)3 (s) 50H- (aq) -0.13 Cu2+ (aq)2e- Cu (s) 0.3419 Cu2+ (aq) e Cu (aq) 0.153 Cu (aq)e-Cu (s) 0.521 F2 (g)2e 2F (aq) Fe2+ (aq)2e 2.866 Fe (s) -0.447 Fe3+ (aq)e"- Fe2+ (aq) 0.771 Fe3+ (aq) 3e Fe (s) -0.037 2H+ (aq) 2e H2 (g) 0.000 2H20 ()2e- H2 (g) + 20H (aq) -0.8277

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Expert Answer

Step 1

The reaction is written as follows:

5Zn(s) + 2IO3- (aq) + 12H+(aq) →5Zn2+(aq) + I2(s) + 6H2O(l)

 

The oxidation half reaction is shown in equation (1).

The reduction half reaction is shown in equation (2).

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Zn(s) Zn2 (aq) 2e E^ -0.7618 V ...... (1) anode 210, (aq)12H (aq) +10e ->I, (s)+6H20(1) Ehode 1.195 V ...... (2)

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Step 2

The standard cell potential is calculated as follows:

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EoE -E cathode cell anode 1.195 V - (-0.7618 V) 1.9568 V

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Step 3

The overall reaction is written by combining equation (1) and (2) and multiplying equation (1) by ...

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5Zn (s)>5Zn2* (aq) + 10e (3) 210, (aq)+12H (aq)+10e -I, (s)+6H20(1) SZn(s) +210 (aq)+12H* (aq) -> 5Zn2 (aq) +I, (s)+6H20(1) AG=-nFEl =-10 molx 96485 J/V.molx1.9568 V ...... (4) -1888018.48 J

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