NbBH4 how many hydrides available for reduction?

[Arman Bonakdarpour]

Hello,

I am using NaBH4 to reduce cations (M3+). Would it take 3 NaBH4 to
reduce it (each giving one hydride) or are all 4 H available for
reduction of the cations?

Best Regards,
Arman

[Ed Ferris]

Post by Arman Bonakdarpour
Hello,
I am using NaBH4 to reduce cations (M3+). Would it take 3 NaBH4 to
reduce it (each giving one hydride) or are all 4 H available for
reduction of the cations?
Best Regards,
Arman

According to the CRC Handbook, each NaBH4 takes up EIGHT electrons:
H2BO3- + 5H2O + 8e- = BH4- + 8OH- -1.24 V
so reducing, say, Fe+3 to Fe requires 3/8ths mole BH4 per mole Fe ion.
(H2BO3- = H2O + BO2-. I'm told you get the BO2- instead of borate.)