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Perhatikan data potensial reduksi standar berikut! Au 3 + + 3 e − Cl 2 ​ + 2 e − Sn 2 + + 2 e − Na + + e − ​ → → → → ​ Au E 0 = + 1 , 42 V 2 Cl − E 0 = + 1 , 36 V Sn E 0 = − 0 , 14 V Na E 0 = − 2 , 71 V ​ Reaksi berikut yang berlangsung tidak spontan adalah ....

Perhatikan data potensial reduksi standar berikut!
 

    
 

Reaksi berikut yang berlangsung tidak spontan adalah ....

  1. begin mathsize 14px style Na and Au to the power of 3 plus sign yields Na to the power of plus sign and Au end style 

  2. begin mathsize 14px style Na and Sn to the power of 2 plus sign yields Na to the power of plus sign and Sn end style 

  3. begin mathsize 14px style Cl to the power of minus sign and Sn to the power of 2 plus sign yields Cl subscript 2 and Sn end style 

  4. begin mathsize 14px style Cl to the power of minus sign and Au to the power of 3 plus sign yields Cl subscript 2 and Au end style 

  5. begin mathsize 14px style Sn and Au to the power of 3 plus sign yields Sn to the power of 2 plus sign and Au end style 

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

jawaban yang tepat adalah C.

Pembahasan

Opsi A: Opsi B: Opsi C: E 0 sel ​ E 0 sel ​ E 0 sel ​ ​ = = = ​ E 0 reduksi ​ − E 0 oksidasi ​ ( − 0 , 14 V ) − ( + 1 , 36 V ) − 1 , 50 V ( tidak spontan ) ​ Opsi D: Opsi E: Jadi, jawaban yang tepat adalah C.

Opsi A:
 

 
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell E to the power of 0 subscript sel end cell equals cell E to the power of 0 subscript reduksi minus sign E to the power of 0 subscript oksidasi end cell row cell E to the power of 0 subscript sel end cell equals cell plus 1 comma 42 space V minus sign left parenthesis minus sign 2 comma 71 space V right parenthesis end cell row cell E to the power of 0 subscript sel end cell equals cell plus 4 comma 13 space V space open parentheses spontan close parentheses end cell end table end style   
 

Opsi B: 
 

 
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell E to the power of 0 subscript sel end cell equals cell E to the power of 0 subscript reduksi minus sign E to the power of 0 subscript oksidasi end cell row cell E to the power of 0 subscript sel end cell equals cell left parenthesis minus sign 0 comma 14 space V right parenthesis minus sign left parenthesis minus sign 2 comma 71 space V right parenthesis end cell row cell E to the power of 0 subscript sel end cell equals cell plus 2 comma 57 space V space open parentheses spontan close parentheses end cell end table end style    
 

Opsi C: 
 

 
 

        
 

Opsi D:
 

 
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell E to the power of 0 subscript sel end cell equals cell E to the power of 0 subscript reduksi minus sign E to the power of 0 subscript oksidasi end cell row cell E to the power of 0 subscript sel end cell equals cell plus 1 comma 42 V minus sign left parenthesis plus 1 comma 36 space V right parenthesis end cell row cell E to the power of 0 subscript sel end cell equals cell plus 0 comma 06 space V space open parentheses spontan close parentheses end cell end table end style   
 

Opsi E:
 

 
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell E to the power of 0 subscript sel end cell equals cell E to the power of 0 subscript reduksi minus sign E to the power of 0 subscript oksidasi end cell row cell E to the power of 0 subscript sel end cell equals cell plus 1 comma 42 V minus sign left parenthesis minus sign 0 comma 14 space V right parenthesis end cell row cell E to the power of 0 subscript sel end cell equals cell plus 1 comma 56 space V space open parentheses spontan close parentheses end cell end table end style   
 

Jadi, jawaban yang tepat adalah C.

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Perhatikan data potensial reduksi standar berikut! Au 3 + + 3 e − → Au E 0 = + 1 , 42 V Cl 2 ​ + 2 e − → 2 Cl − E 0 = + 1 , 36 V Sn 2 + + 2 e − → Sn E 0 = − 0 , 14 V Na + + e − → Na E 0 = − 2 , 71 ...

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