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1840 United States presidential election in Alabama

From Wikipedia, the free encyclopedia

1840 United States presidential election in Alabama

← 1836 October 30 – December 2, 1840 1844 →
Van Buren.jpg
William Henry Harrison (cropped).jpg
Nominee Martin Van Buren William Henry Harrison
Party Democratic Whig
Home state New York Ohio
Running mate none John Tyler
Electoral vote 7 0
Popular vote 33,996 28,518
Percentage 54.38% 45.62%

President before election

Martin Van Buren

Elected President

William Henry Harrison

The 1840 United States presidential election in Alabama took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

Alabama voted for the Democratic candidate, Martin Van Buren, over Whig candidate William Henry Harrison. Van Buren won Alabama by a margin of 8.76%. This is the last time that Alabama did not vote the same as neighboring Mississippi.

YouTube Encyclopedic

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  • The Election of 1844 Explained
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  • The 1832 Election Explained
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1840 United States presidential election in Alabama[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 33,996 54.38% 7 100.00%
Whig William Henry Harrison of Ohio John Tyler of Virginia 28,515 45.62% 0 0.00%
Total 62,511 100.00% 7 100.00%


  1. ^ "1840 Presidential General Election Results - Alabama". U.S. Election Atlas. Retrieved December 23, 2013.

This page was last edited on 24 February 2021, at 12:22
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