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1840 United States presidential election in Maine

From Wikipedia, the free encyclopedia

1840 United States presidential election in Maine
Flag of the United States (1837–1845).svg

← 1836 October 30 - December 2, 1840 1844 →
William Henry Harrison (cropped).jpg
Martin Van Buren MET ap93.19.2.jpg
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 10 0
Popular vote 46,612 46,190
Percentage 50.23% 49.77%

President before election

Martin Van Buren

Elected President

William Henry Harrison

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

Maine voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won the state by a very narrow margin of 0.46%.

Maine was typically a Democratic state during the Second Party System, however, with Harrison narrowly winning the state, this would be the only time that a Whig presidential candidate would win Maine.


1840 United States presidential election in Maine[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio John Tyler of Virginia 46,612 50.23% 10 100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 46,190 49.77% 0 0.00%
Total 92,802 100.00% 10 100.00%

See also


  1. ^ "1840 Presidential General Election Results - Maine". U.S. Election Atlas. Retrieved 23 December 2013.
This page was last edited on 21 April 2021, at 22:04
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